Is it possible to output the numpad's pattern using loops?
I just started learning C (coding in general) a few months ago. Earlier today when I was in class, I looked at the numpad and wondered whether I would be able to replicate the pattern using nested loops in C.
7 8 9
4 5 6
1 2 3 // This pattern.
I tried to do it myself for a bit, using for loops primarily. Thanks for any help.
#include<stdio.h>
int main()
{
int row, col, i;
printf("Up to what integer? ");
scanf("%d", &row);
for(i=1; i<=row; i++)
{
for(col=1; col<=10; col++)
{
printf(" %d ", i*col);
}
printf("n");
}
}
Edit: Added supplementary code. Something like this, except to print 3 rows and 3 columns.
c design-patterns printing nested-loops
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
add a comment |
I just started learning C (coding in general) a few months ago. Earlier today when I was in class, I looked at the numpad and wondered whether I would be able to replicate the pattern using nested loops in C.
7 8 9
4 5 6
1 2 3 // This pattern.
I tried to do it myself for a bit, using for loops primarily. Thanks for any help.
#include<stdio.h>
int main()
{
int row, col, i;
printf("Up to what integer? ");
scanf("%d", &row);
for(i=1; i<=row; i++)
{
for(col=1; col<=10; col++)
{
printf(" %d ", i*col);
}
printf("n");
}
}
Edit: Added supplementary code. Something like this, except to print 3 rows and 3 columns.
c design-patterns printing nested-loops
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
Sure it is possible. What have you tried so far and were are you stuck?
– Gerhardh
Nov 22 '18 at 11:12
I've tried with integers i, x, y; with for loops i =1. I've tried with the base number being 9 and 1.
– partynextdoor18
Nov 22 '18 at 11:16
1
Please show a MCVE. Without code its pointless to guess that your problem is. Also include the wrong output you got.
– Gerhardh
Nov 22 '18 at 11:17
add a comment |
I just started learning C (coding in general) a few months ago. Earlier today when I was in class, I looked at the numpad and wondered whether I would be able to replicate the pattern using nested loops in C.
7 8 9
4 5 6
1 2 3 // This pattern.
I tried to do it myself for a bit, using for loops primarily. Thanks for any help.
#include<stdio.h>
int main()
{
int row, col, i;
printf("Up to what integer? ");
scanf("%d", &row);
for(i=1; i<=row; i++)
{
for(col=1; col<=10; col++)
{
printf(" %d ", i*col);
}
printf("n");
}
}
Edit: Added supplementary code. Something like this, except to print 3 rows and 3 columns.
c design-patterns printing nested-loops
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
I just started learning C (coding in general) a few months ago. Earlier today when I was in class, I looked at the numpad and wondered whether I would be able to replicate the pattern using nested loops in C.
7 8 9
4 5 6
1 2 3 // This pattern.
I tried to do it myself for a bit, using for loops primarily. Thanks for any help.
#include<stdio.h>
int main()
{
int row, col, i;
printf("Up to what integer? ");
scanf("%d", &row);
for(i=1; i<=row; i++)
{
for(col=1; col<=10; col++)
{
printf(" %d ", i*col);
}
printf("n");
}
}
Edit: Added supplementary code. Something like this, except to print 3 rows and 3 columns.
c design-patterns printing nested-loops
c design-patterns printing nested-loops
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
edited Nov 22 '18 at 11:58
partynextdoor18
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
asked Nov 22 '18 at 11:05
partynextdoor18partynextdoor18
175
175
Sure it is possible. What have you tried so far and were are you stuck?
– Gerhardh
Nov 22 '18 at 11:12
I've tried with integers i, x, y; with for loops i =1. I've tried with the base number being 9 and 1.
– partynextdoor18
Nov 22 '18 at 11:16
1
Please show a MCVE. Without code its pointless to guess that your problem is. Also include the wrong output you got.
– Gerhardh
Nov 22 '18 at 11:17
add a comment |
Sure it is possible. What have you tried so far and were are you stuck?
– Gerhardh
Nov 22 '18 at 11:12
I've tried with integers i, x, y; with for loops i =1. I've tried with the base number being 9 and 1.
– partynextdoor18
Nov 22 '18 at 11:16
1
Please show a MCVE. Without code its pointless to guess that your problem is. Also include the wrong output you got.
– Gerhardh
Nov 22 '18 at 11:17
Sure it is possible. What have you tried so far and were are you stuck?
– Gerhardh
Nov 22 '18 at 11:12
Sure it is possible. What have you tried so far and were are you stuck?
– Gerhardh
Nov 22 '18 at 11:12
I've tried with integers i, x, y; with for loops i =1. I've tried with the base number being 9 and 1.
– partynextdoor18
Nov 22 '18 at 11:16
I've tried with integers i, x, y; with for loops i =1. I've tried with the base number being 9 and 1.
– partynextdoor18
Nov 22 '18 at 11:16
1
1
Please show a MCVE. Without code its pointless to guess that your problem is. Also include the wrong output you got.
– Gerhardh
Nov 22 '18 at 11:17
Please show a MCVE. Without code its pointless to guess that your problem is. Also include the wrong output you got.
– Gerhardh
Nov 22 '18 at 11:17
add a comment |
2 Answers
2
active
oldest
votes
The numpad pattern has the equation 3*i + j with i going from 2 to 0 and j going from 1 to 3.
So use these values as upper and lower limits of i and j in nested for loops.
#include <stdio.h>
int main(){
for(int i = 2; i >= 0; i--){
for(int j = 1; j <= 3; j++)
printf("%d ", 3 * i + j);
printf("n");
}
return 0;
}
See it live here.
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
add a comment |
Here's how you can do it:
for(int i = 0; i < 3; ++i){
for(int j = 3; j > 0; --j)
printf("%d ", (10 - j) - i * 3);
printf("n");
}
answered Nov 22 '18 at 11:38
sstefansstefan
336311
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The numpad pattern has the equation 3*i + j with i going from 2 to 0 and j going from 1 to 3.
So use these values as upper and lower limits of i and j in nested for loops.
#include <stdio.h>
int main(){
for(int i = 2; i >= 0; i--){
for(int j = 1; j <= 3; j++)
printf("%d ", 3 * i + j);
printf("n");
}
return 0;
}
See it live here.
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
add a comment |
The numpad pattern has the equation 3*i + j with i going from 2 to 0 and j going from 1 to 3.
So use these values as upper and lower limits of i and j in nested for loops.
#include <stdio.h>
int main(){
for(int i = 2; i >= 0; i--){
for(int j = 1; j <= 3; j++)
printf("%d ", 3 * i + j);
printf("n");
}
return 0;
}
See it live here.
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
add a comment |
The numpad pattern has the equation 3*i + j with i going from 2 to 0 and j going from 1 to 3.
So use these values as upper and lower limits of i and j in nested for loops.
#include <stdio.h>
int main(){
for(int i = 2; i >= 0; i--){
for(int j = 1; j <= 3; j++)
printf("%d ", 3 * i + j);
printf("n");
}
return 0;
}
See it live here.
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
The numpad pattern has the equation 3*i + j with i going from 2 to 0 and j going from 1 to 3.
So use these values as upper and lower limits of i and j in nested for loops.
#include <stdio.h>
int main(){
for(int i = 2; i >= 0; i--){
for(int j = 1; j <= 3; j++)
printf("%d ", 3 * i + j);
printf("n");
}
return 0;
}
See it live here.
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
edited Nov 22 '18 at 12:04
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
answered Nov 22 '18 at 11:57
P.WP.W
16.8k41555
16.8k41555
add a comment |
add a comment |
Here's how you can do it:
for(int i = 0; i < 3; ++i){
for(int j = 3; j > 0; --j)
printf("%d ", (10 - j) - i * 3);
printf("n");
}
answered Nov 22 '18 at 11:38
sstefansstefan
336311
add a comment |
Here's how you can do it:
for(int i = 0; i < 3; ++i){
for(int j = 3; j > 0; --j)
printf("%d ", (10 - j) - i * 3);
printf("n");
}
answered Nov 22 '18 at 11:38
sstefansstefan
336311
add a comment |
Here's how you can do it:
for(int i = 0; i < 3; ++i){
for(int j = 3; j > 0; --j)
printf("%d ", (10 - j) - i * 3);
printf("n");
}
answered Nov 22 '18 at 11:38
sstefansstefan
336311
Here's how you can do it:
for(int i = 0; i < 3; ++i){
for(int j = 3; j > 0; --j)
printf("%d ", (10 - j) - i * 3);
printf("n");
}
answered Nov 22 '18 at 11:38
sstefansstefan
336311
answered Nov 22 '18 at 11:38
sstefansstefan
336311
answered Nov 22 '18 at 11:38
sstefansstefan
336311
answered Nov 22 '18 at 11:38
sstefansstefan
336311
336311
add a comment |
add a comment |
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Sure it is possible. What have you tried so far and were are you stuck?
– Gerhardh
Nov 22 '18 at 11:12
I've tried with integers i, x, y; with for loops i =1. I've tried with the base number being 9 and 1.
– partynextdoor18
Nov 22 '18 at 11:16
1
Please show a MCVE. Without code its pointless to guess that your problem is. Also include the wrong output you got.
– Gerhardh
Nov 22 '18 at 11:17