How to restart main() from the beginning on an error condition?
1
#include <stdio.h>
int main()
{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
/* want to restart main() again here */
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if(num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
if (ask == 'y' || ask == 'Y') {
printf("n");
main();
}
else {
printf("Thank you for using the program. Please give full marks.");
}
}
return 0;
}
c
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
|
show 6 more comments
1
#include <stdio.h>
int main()
{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
/* want to restart main() again here */
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if(num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
if (ask == 'y' || ask == 'Y') {
printf("n");
main();
}
else {
printf("Thank you for using the program. Please give full marks.");
}
}
return 0;
}
c
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
2
Welcome to stackoverflow.com. Please take some time to read the help pages, especially the sections named "What topics can I ask about here?" and "What types of questions should I avoid asking?". Also please take the tour and read about how to ask good questions. Lastly please read this question checklist.
– Some programmer dude
Nov 19 '18 at 10:27
2
As for "repeating" something, use loops.
– Some programmer dude
Nov 19 '18 at 10:28
Possible duplicate of calling main() in main() in c
– borrible
Nov 19 '18 at 10:31
3
@John Hunter. Are you aware of loops in programming? Also, are you aware of recursion? Because it seems you are trying to use recursion for something that a simple loop would be more than enough
– kyriakosSt
Nov 19 '18 at 10:35
2
@JohnHunter I understand what you need to do, but I need to know if you are aware of loops in the first place in order to explain. Kristjan Kica 's answer however seems pretty good
– kyriakosSt
Nov 19 '18 at 10:59
|
show 6 more comments
1
1
1
#include <stdio.h>
int main()
{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
/* want to restart main() again here */
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if(num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
if (ask == 'y' || ask == 'Y') {
printf("n");
main();
}
else {
printf("Thank you for using the program. Please give full marks.");
}
}
return 0;
}
c
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
#include <stdio.h>
int main()
{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
/* want to restart main() again here */
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if(num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
if (ask == 'y' || ask == 'Y') {
printf("n");
main();
}
else {
printf("Thank you for using the program. Please give full marks.");
}
}
return 0;
}
c
c
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
edited Nov 20 '18 at 16:36
HostileFork
25.6k778133
25.6k778133
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
asked Nov 19 '18 at 10:26
John HunterJohn Hunter
134
134
2
Welcome to stackoverflow.com. Please take some time to read the help pages, especially the sections named "What topics can I ask about here?" and "What types of questions should I avoid asking?". Also please take the tour and read about how to ask good questions. Lastly please read this question checklist.
– Some programmer dude
Nov 19 '18 at 10:27
2
As for "repeating" something, use loops.
– Some programmer dude
Nov 19 '18 at 10:28
Possible duplicate of calling main() in main() in c
– borrible
Nov 19 '18 at 10:31
3
@John Hunter. Are you aware of loops in programming? Also, are you aware of recursion? Because it seems you are trying to use recursion for something that a simple loop would be more than enough
– kyriakosSt
Nov 19 '18 at 10:35
2
@JohnHunter I understand what you need to do, but I need to know if you are aware of loops in the first place in order to explain. Kristjan Kica 's answer however seems pretty good
– kyriakosSt
Nov 19 '18 at 10:59
|
show 6 more comments
2
Welcome to stackoverflow.com. Please take some time to read the help pages, especially the sections named "What topics can I ask about here?" and "What types of questions should I avoid asking?". Also please take the tour and read about how to ask good questions. Lastly please read this question checklist.
– Some programmer dude
Nov 19 '18 at 10:27
2
As for "repeating" something, use loops.
– Some programmer dude
Nov 19 '18 at 10:28
Possible duplicate of calling main() in main() in c
– borrible
Nov 19 '18 at 10:31
3
@John Hunter. Are you aware of loops in programming? Also, are you aware of recursion? Because it seems you are trying to use recursion for something that a simple loop would be more than enough
– kyriakosSt
Nov 19 '18 at 10:35
2
@JohnHunter I understand what you need to do, but I need to know if you are aware of loops in the first place in order to explain. Kristjan Kica 's answer however seems pretty good
– kyriakosSt
Nov 19 '18 at 10:59
2
2
Welcome to stackoverflow.com. Please take some time to read the help pages, especially the sections named "What topics can I ask about here?" and "What types of questions should I avoid asking?". Also please take the tour and read about how to ask good questions. Lastly please read this question checklist.
– Some programmer dude
Nov 19 '18 at 10:27
Welcome to stackoverflow.com. Please take some time to read the help pages, especially the sections named "What topics can I ask about here?" and "What types of questions should I avoid asking?". Also please take the tour and read about how to ask good questions. Lastly please read this question checklist.
– Some programmer dude
Nov 19 '18 at 10:27
2
2
As for "repeating" something, use loops.
– Some programmer dude
Nov 19 '18 at 10:28
As for "repeating" something, use loops.
– Some programmer dude
Nov 19 '18 at 10:28
Possible duplicate of calling main() in main() in c
– borrible
Nov 19 '18 at 10:31
Possible duplicate of calling main() in main() in c
– borrible
Nov 19 '18 at 10:31
3
3
@John Hunter. Are you aware of loops in programming? Also, are you aware of recursion? Because it seems you are trying to use recursion for something that a simple loop would be more than enough
– kyriakosSt
Nov 19 '18 at 10:35
@John Hunter. Are you aware of loops in programming? Also, are you aware of recursion? Because it seems you are trying to use recursion for something that a simple loop would be more than enough
– kyriakosSt
Nov 19 '18 at 10:35
2
2
@JohnHunter I understand what you need to do, but I need to know if you are aware of loops in the first place in order to explain. Kristjan Kica 's answer however seems pretty good
– kyriakosSt
Nov 19 '18 at 10:59
@JohnHunter I understand what you need to do, but I need to know if you are aware of loops in the first place in order to explain. Kristjan Kica 's answer however seems pretty good
– kyriakosSt
Nov 19 '18 at 10:59
|
show 6 more comments
3 Answers
3
active
oldest
votes
7
To answer your question.
I would not recommend calling main.
You could create another function that has all you code.
Inside main, you call that function.
You can call a function inside that function (called recursion)
However, a simple loop could do the job.
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
}
Edit: To answer the comment to this question what you want to do is:
while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
And remove the else
EDIT2: Full code. Note that the expression cannot be more than 99 characters.
#include <stdio.h>
int main()
{
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
char string[100];
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
while(1){
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
printf("nInvalid input! Please try again...nn");
else
break;
}
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
return 0;
}
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
|
show 9 more comments
0
@Kristjan Kica answer is good. I think you are using spaces in your input like 1 + 2.
According to manual page of scanf
All conversions are introduced by the % (percent sign) character. The format string may also contain other characters. White space
(such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else
matches only itself. Scanning stops when an input character does not match such a format character. Scanning also stops when an input
conversion cannot be made.
Remove the spaces and try again.
Example:
1+2 should work by the changes mentioned in kristijan answer.
Edit:
Replace the line in @Kristjan Kica answer
while(ask == 'y' || ask == 'Y') ;
with
}while(ask == 'y' || ask == 'Y') ;
Edit 2:
Last closing } should be your main functions closing brace.
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
add a comment |
0
Finally solved it. All thanks to Kristjan Kica...
#include <stdio.h>
int main(void)
{
float num1;
float num2;
float ans;
char symbol;
char ask;
char string[100];
fflush(stdin);
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by
Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
{
printf("nInvalid input! Please try again...nn");
main();
}
else
{
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
if (ask == 'y' || ask == 'Y')
{
main();
}
else {
return 0;
}
}
}
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
To answer your question.
I would not recommend calling main.
You could create another function that has all you code.
Inside main, you call that function.
You can call a function inside that function (called recursion)
However, a simple loop could do the job.
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
}
Edit: To answer the comment to this question what you want to do is:
while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
And remove the else
EDIT2: Full code. Note that the expression cannot be more than 99 characters.
#include <stdio.h>
int main()
{
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
char string[100];
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
while(1){
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
printf("nInvalid input! Please try again...nn");
else
break;
}
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
return 0;
}
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
|
show 9 more comments
7
To answer your question.
I would not recommend calling main.
You could create another function that has all you code.
Inside main, you call that function.
You can call a function inside that function (called recursion)
However, a simple loop could do the job.
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
}
Edit: To answer the comment to this question what you want to do is:
while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
And remove the else
EDIT2: Full code. Note that the expression cannot be more than 99 characters.
#include <stdio.h>
int main()
{
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
char string[100];
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
while(1){
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
printf("nInvalid input! Please try again...nn");
else
break;
}
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
return 0;
}
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
|
show 9 more comments
7
7
7
To answer your question.
I would not recommend calling main.
You could create another function that has all you code.
Inside main, you call that function.
You can call a function inside that function (called recursion)
However, a simple loop could do the job.
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
}
Edit: To answer the comment to this question what you want to do is:
while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
And remove the else
EDIT2: Full code. Note that the expression cannot be more than 99 characters.
#include <stdio.h>
int main()
{
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
char string[100];
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
while(1){
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
printf("nInvalid input! Please try again...nn");
else
break;
}
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
return 0;
}
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
To answer your question.
I would not recommend calling main.
You could create another function that has all you code.
Inside main, you call that function.
You can call a function inside that function (called recursion)
However, a simple loop could do the job.
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
printf("Please type the expression you want to calculate: ");
if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
else {
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
}
Edit: To answer the comment to this question what you want to do is:
while(scanf("%f%1s%f",&num1,&symbol,&num2) != 3)
{
printf("nInvalid input! Please try again...nn");
}
And remove the else
EDIT2: Full code. Note that the expression cannot be more than 99 characters.
#include <stdio.h>
int main()
{
float num1;
float num2;
float ans = 0.0;
char symbol;
char ask;
char string[100];
do{
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
while(1){
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
printf("nInvalid input! Please try again...nn");
else
break;
}
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
}while(ask == 'y' || ask == 'Y') ;
printf("Thank you for using the program. Please give full marks.");
return 0;
}
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
edited Nov 20 '18 at 12:36
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
answered Nov 19 '18 at 10:32
Kristjan KicaKristjan Kica
2,2071926
2,2071926
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
|
show 9 more comments
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
Hi, please note that I was referring to this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 10:37
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
You have called main elsewhere. Let me edit my answer. The same principle applies
– Kristjan Kica
Nov 19 '18 at 10:42
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
Showing invalid input infinite no. times... Something like this... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again.... Invalid input. Please try again....
– John Hunter
Nov 19 '18 at 10:50
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
If scanf can't find matching input for the expected format, it will return. With this loop you no longer know where the input is, and will probably fail indefinitely. This pattern is generally a bad idea. Consider using line based input with fgets and then sscanf.
– paddy
Nov 19 '18 at 11:36
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
Please give an example @paddy... I am really new to programming....
– John Hunter
Nov 19 '18 at 11:58
|
show 9 more comments
0
@Kristjan Kica answer is good. I think you are using spaces in your input like 1 + 2.
According to manual page of scanf
All conversions are introduced by the % (percent sign) character. The format string may also contain other characters. White space
(such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else
matches only itself. Scanning stops when an input character does not match such a format character. Scanning also stops when an input
conversion cannot be made.
Remove the spaces and try again.
Example:
1+2 should work by the changes mentioned in kristijan answer.
Edit:
Replace the line in @Kristjan Kica answer
while(ask == 'y' || ask == 'Y') ;
with
}while(ask == 'y' || ask == 'Y') ;
Edit 2:
Last closing } should be your main functions closing brace.
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
add a comment |
0
@Kristjan Kica answer is good. I think you are using spaces in your input like 1 + 2.
According to manual page of scanf
All conversions are introduced by the % (percent sign) character. The format string may also contain other characters. White space
(such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else
matches only itself. Scanning stops when an input character does not match such a format character. Scanning also stops when an input
conversion cannot be made.
Remove the spaces and try again.
Example:
1+2 should work by the changes mentioned in kristijan answer.
Edit:
Replace the line in @Kristjan Kica answer
while(ask == 'y' || ask == 'Y') ;
with
}while(ask == 'y' || ask == 'Y') ;
Edit 2:
Last closing } should be your main functions closing brace.
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
add a comment |
0
0
0
@Kristjan Kica answer is good. I think you are using spaces in your input like 1 + 2.
According to manual page of scanf
All conversions are introduced by the % (percent sign) character. The format string may also contain other characters. White space
(such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else
matches only itself. Scanning stops when an input character does not match such a format character. Scanning also stops when an input
conversion cannot be made.
Remove the spaces and try again.
Example:
1+2 should work by the changes mentioned in kristijan answer.
Edit:
Replace the line in @Kristjan Kica answer
while(ask == 'y' || ask == 'Y') ;
with
}while(ask == 'y' || ask == 'Y') ;
Edit 2:
Last closing } should be your main functions closing brace.
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
@Kristjan Kica answer is good. I think you are using spaces in your input like 1 + 2.
According to manual page of scanf
All conversions are introduced by the % (percent sign) character. The format string may also contain other characters. White space
(such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input. Everything else
matches only itself. Scanning stops when an input character does not match such a format character. Scanning also stops when an input
conversion cannot be made.
Remove the spaces and try again.
Example:
1+2 should work by the changes mentioned in kristijan answer.
Edit:
Replace the line in @Kristjan Kica answer
while(ask == 'y' || ask == 'Y') ;
with
}while(ask == 'y' || ask == 'Y') ;
Edit 2:
Last closing } should be your main functions closing brace.
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
edited Nov 19 '18 at 13:12
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
answered Nov 19 '18 at 12:24
bhanu7kbhanu7k
536
536
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
add a comment |
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Upon running Kristjan's code, it say expected while before }
– John Hunter
Nov 19 '18 at 12:50
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Yup it's working... But how can I do the same thing in this line... if(scanf("%f%1s%f",&num1,&symbol,&num2) != 3) { printf("nInvalid input! Please try again...nn"); }
– John Hunter
Nov 19 '18 at 13:23
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
Are u getting the output or is it just compiled?
– bhanu7k
Nov 19 '18 at 13:29
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
You can achieve the same with if statement(remove while inside do-while and try again).
– bhanu7k
Nov 19 '18 at 13:31
add a comment |
0
Finally solved it. All thanks to Kristjan Kica...
#include <stdio.h>
int main(void)
{
float num1;
float num2;
float ans;
char symbol;
char ask;
char string[100];
fflush(stdin);
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by
Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
{
printf("nInvalid input! Please try again...nn");
main();
}
else
{
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
if (ask == 'y' || ask == 'Y')
{
main();
}
else {
return 0;
}
}
}
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
add a comment |
0
Finally solved it. All thanks to Kristjan Kica...
#include <stdio.h>
int main(void)
{
float num1;
float num2;
float ans;
char symbol;
char ask;
char string[100];
fflush(stdin);
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by
Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
{
printf("nInvalid input! Please try again...nn");
main();
}
else
{
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
if (ask == 'y' || ask == 'Y')
{
main();
}
else {
return 0;
}
}
}
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
add a comment |
0
0
0
Finally solved it. All thanks to Kristjan Kica...
#include <stdio.h>
int main(void)
{
float num1;
float num2;
float ans;
char symbol;
char ask;
char string[100];
fflush(stdin);
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by
Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
{
printf("nInvalid input! Please try again...nn");
main();
}
else
{
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
if (ask == 'y' || ask == 'Y')
{
main();
}
else {
return 0;
}
}
}
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
Finally solved it. All thanks to Kristjan Kica...
#include <stdio.h>
int main(void)
{
float num1;
float num2;
float ans;
char symbol;
char ask;
char string[100];
fflush(stdin);
printf("Hi!nWelcome!nThis is an expression based calculatorndeveloped by
Sankasuvra Bhattacharyan");
printf("that performs arithmetic operations onntwo numbers.n");
printf("Please type the expression you want to calculate: ");
fgets (string , 100 ,stdin);
if(sscanf( string, "%f%1s%f",&num1,&symbol,&num2)!=3)
{
printf("nInvalid input! Please try again...nn");
main();
}
else
{
switch(symbol) {
case '+' : ans = num1 + num2;
break;
case '-' : ans = num1 - num2;
break;
case '*' :
case 'x' :
ans = num1 * num2;
break;
case '/' :
if (num2 == 0) {
printf("Division by zero is not possible!nPlease try again...nn");
return main();
}
else {
ans = num1 / num2;
break;
}
default :
printf("nInvalid input! Please try again...nn");
return main();
}
printf("The answer is %gn",ans);
printf("nTo use the calculator again, type 'Y'. ");
printf("To exit, type any other character...n");
scanf("%s",&ask);
printf("n");
if (ask == 'y' || ask == 'Y')
{
main();
}
else {
return 0;
}
}
}
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
answered Nov 20 '18 at 16:26
John HunterJohn Hunter
134
134
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– Some programmer dude
Nov 19 '18 at 10:27
2
As for "repeating" something, use loops.
– Some programmer dude
Nov 19 '18 at 10:28
Possible duplicate of calling main() in main() in c
– borrible
Nov 19 '18 at 10:31
3
@John Hunter. Are you aware of loops in programming? Also, are you aware of recursion? Because it seems you are trying to use recursion for something that a simple loop would be more than enough
– kyriakosSt
Nov 19 '18 at 10:35
2
@JohnHunter I understand what you need to do, but I need to know if you are aware of loops in the first place in order to explain. Kristjan Kica 's answer however seems pretty good
– kyriakosSt
Nov 19 '18 at 10:59