can you explain me why it is possible p[-1]?
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int *p[10]={5,663,36,6};
*(p - 1) = 'e';
int c=*(p-1);
printf("%c",c);
i am not able to understand why we use negative number in array index
*(p - 1) = 'e';
c
edited Nov 23 '18 at 15:21
Blaze
7,5241832
asked Nov 23 '18 at 15:13
Learning cLearning c
6
add a comment |
int *p[10]={5,663,36,6};
*(p - 1) = 'e';
int c=*(p-1);
printf("%c",c);
i am not able to understand why we use negative number in array index
*(p - 1) = 'e';
c
edited Nov 23 '18 at 15:21
Blaze
7,5241832
asked Nov 23 '18 at 15:13
Learning cLearning c
6
5
You can't, really. It's undefined behavior, so there's no telling what happens. It might work as intended, but it might just as well crash the program or do something even worse. In practice what most likely happens is that it calculates the memory address and just writes/reads to/from it anyway, which may cause data corrupion (because it's overwriting something else), stack corruption (by writing over the stack pointer), or make the OS detect an illegal memory access and crash the program.
– Blaze
Nov 23 '18 at 15:15
Also, you can't assign an integer value to a pointer without a cast.
– EOF
Nov 23 '18 at 15:17
add a comment |
int *p[10]={5,663,36,6};
*(p - 1) = 'e';
int c=*(p-1);
printf("%c",c);
i am not able to understand why we use negative number in array index
*(p - 1) = 'e';
c
edited Nov 23 '18 at 15:21
Blaze
7,5241832
asked Nov 23 '18 at 15:13
Learning cLearning c
6
int *p[10]={5,663,36,6};
*(p - 1) = 'e';
int c=*(p-1);
printf("%c",c);
i am not able to understand why we use negative number in array index
*(p - 1) = 'e';
c
c
edited Nov 23 '18 at 15:21
Blaze
7,5241832
asked Nov 23 '18 at 15:13
Learning cLearning c
6
edited Nov 23 '18 at 15:21
Blaze
7,5241832
asked Nov 23 '18 at 15:13
Learning cLearning c
6
edited Nov 23 '18 at 15:21
Blaze
7,5241832
edited Nov 23 '18 at 15:21
Blaze
7,5241832
edited Nov 23 '18 at 15:21
Blaze
7,5241832
7,5241832
asked Nov 23 '18 at 15:13
Learning cLearning c
6
asked Nov 23 '18 at 15:13
Learning cLearning c
6
asked Nov 23 '18 at 15:13
Learning cLearning c
6
6
5
You can't, really. It's undefined behavior, so there's no telling what happens. It might work as intended, but it might just as well crash the program or do something even worse. In practice what most likely happens is that it calculates the memory address and just writes/reads to/from it anyway, which may cause data corrupion (because it's overwriting something else), stack corruption (by writing over the stack pointer), or make the OS detect an illegal memory access and crash the program.
– Blaze
Nov 23 '18 at 15:15
Also, you can't assign an integer value to a pointer without a cast.
– EOF
Nov 23 '18 at 15:17
add a comment |
5
You can't, really. It's undefined behavior, so there's no telling what happens. It might work as intended, but it might just as well crash the program or do something even worse. In practice what most likely happens is that it calculates the memory address and just writes/reads to/from it anyway, which may cause data corrupion (because it's overwriting something else), stack corruption (by writing over the stack pointer), or make the OS detect an illegal memory access and crash the program.
– Blaze
Nov 23 '18 at 15:15
Also, you can't assign an integer value to a pointer without a cast.
– EOF
Nov 23 '18 at 15:17
5
5
You can't, really. It's undefined behavior, so there's no telling what happens. It might work as intended, but it might just as well crash the program or do something even worse. In practice what most likely happens is that it calculates the memory address and just writes/reads to/from it anyway, which may cause data corrupion (because it's overwriting something else), stack corruption (by writing over the stack pointer), or make the OS detect an illegal memory access and crash the program.
– Blaze
Nov 23 '18 at 15:15
You can't, really. It's undefined behavior, so there's no telling what happens. It might work as intended, but it might just as well crash the program or do something even worse. In practice what most likely happens is that it calculates the memory address and just writes/reads to/from it anyway, which may cause data corrupion (because it's overwriting something else), stack corruption (by writing over the stack pointer), or make the OS detect an illegal memory access and crash the program.
– Blaze
Nov 23 '18 at 15:15
Also, you can't assign an integer value to a pointer without a cast.
– EOF
Nov 23 '18 at 15:17
Also, you can't assign an integer value to a pointer without a cast.
– EOF
Nov 23 '18 at 15:17
add a comment |
3 Answers
3
active
oldest
votes
For your example it would be undefined behaviour, but there are situations where you might want to use it, notably if your pointer was pointing to somewhere inside an array and you check that you are still inside the bounds of the array.
Like in this example...
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv){
char hello="worlld";
char *p;
for(p=hello;*p!='';p++) {
if ((p>hello) && (p[-1]==p[0])) {
printf("%cn",p[0]);
}
}
return(0);
}
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
add a comment |
The language does not prevent you from using negative numbers in indexing of an array or a pointer. This does not meant that it is always correct. i.e. in your example it would access an array element which is 1 position before the beginning of the array. in other words you access invalid memory addres.
However in the situation like the following, where p1 points to a non-0 element of the array, you can use negative indexes:
int p = {1,2,3,4};
int *p1 = &p[1];
int x = *(p1-1);
int y = p1[-1]; // equivalent to the previous one
In both cases 'x' and 'y' will become '1';
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
add a comment |
i am not able to understand why we use negative number in array index
That's because
you apparently think
is an array operator, but it is not. It is a pointer operator, defined in terms of pointer arithmetic, which, in a general sense, permits subtracting integers from pointers.you seem to have an expectation of some particular kind of behavior arising from evaluating your example code, but it exhibits undefined behavior on account of performing arithmetic on pointer
pthat does not produce a result pointing into (or just past the end of) the same object thatppoints [in]to. "Undefined" means exactly what it says. Although an obvious program failure or error message might be emitted, you cannot rely on that, or on any other particular behavior. Of the entire program.
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
2
Most noticeably one can also write-1[x]reversing the 'array' and 'index' because the standard saysa[b]is just*(a+b)and + is commutative.
– Goswin von Brederlow
Nov 23 '18 at 16:13
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
For your example it would be undefined behaviour, but there are situations where you might want to use it, notably if your pointer was pointing to somewhere inside an array and you check that you are still inside the bounds of the array.
Like in this example...
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv){
char hello="worlld";
char *p;
for(p=hello;*p!='';p++) {
if ((p>hello) && (p[-1]==p[0])) {
printf("%cn",p[0]);
}
}
return(0);
}
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
add a comment |
For your example it would be undefined behaviour, but there are situations where you might want to use it, notably if your pointer was pointing to somewhere inside an array and you check that you are still inside the bounds of the array.
Like in this example...
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv){
char hello="worlld";
char *p;
for(p=hello;*p!='';p++) {
if ((p>hello) && (p[-1]==p[0])) {
printf("%cn",p[0]);
}
}
return(0);
}
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
add a comment |
For your example it would be undefined behaviour, but there are situations where you might want to use it, notably if your pointer was pointing to somewhere inside an array and you check that you are still inside the bounds of the array.
Like in this example...
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv){
char hello="worlld";
char *p;
for(p=hello;*p!='';p++) {
if ((p>hello) && (p[-1]==p[0])) {
printf("%cn",p[0]);
}
}
return(0);
}
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
For your example it would be undefined behaviour, but there are situations where you might want to use it, notably if your pointer was pointing to somewhere inside an array and you check that you are still inside the bounds of the array.
Like in this example...
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv){
char hello="worlld";
char *p;
for(p=hello;*p!='';p++) {
if ((p>hello) && (p[-1]==p[0])) {
printf("%cn",p[0]);
}
}
return(0);
}
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
answered Nov 23 '18 at 15:24
Chris TurnerChris Turner
7,31011118
7,31011118
add a comment |
add a comment |
The language does not prevent you from using negative numbers in indexing of an array or a pointer. This does not meant that it is always correct. i.e. in your example it would access an array element which is 1 position before the beginning of the array. in other words you access invalid memory addres.
However in the situation like the following, where p1 points to a non-0 element of the array, you can use negative indexes:
int p = {1,2,3,4};
int *p1 = &p[1];
int x = *(p1-1);
int y = p1[-1]; // equivalent to the previous one
In both cases 'x' and 'y' will become '1';
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
add a comment |
The language does not prevent you from using negative numbers in indexing of an array or a pointer. This does not meant that it is always correct. i.e. in your example it would access an array element which is 1 position before the beginning of the array. in other words you access invalid memory addres.
However in the situation like the following, where p1 points to a non-0 element of the array, you can use negative indexes:
int p = {1,2,3,4};
int *p1 = &p[1];
int x = *(p1-1);
int y = p1[-1]; // equivalent to the previous one
In both cases 'x' and 'y' will become '1';
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
add a comment |
The language does not prevent you from using negative numbers in indexing of an array or a pointer. This does not meant that it is always correct. i.e. in your example it would access an array element which is 1 position before the beginning of the array. in other words you access invalid memory addres.
However in the situation like the following, where p1 points to a non-0 element of the array, you can use negative indexes:
int p = {1,2,3,4};
int *p1 = &p[1];
int x = *(p1-1);
int y = p1[-1]; // equivalent to the previous one
In both cases 'x' and 'y' will become '1';
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
The language does not prevent you from using negative numbers in indexing of an array or a pointer. This does not meant that it is always correct. i.e. in your example it would access an array element which is 1 position before the beginning of the array. in other words you access invalid memory addres.
However in the situation like the following, where p1 points to a non-0 element of the array, you can use negative indexes:
int p = {1,2,3,4};
int *p1 = &p[1];
int x = *(p1-1);
int y = p1[-1]; // equivalent to the previous one
In both cases 'x' and 'y' will become '1';
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
answered Nov 23 '18 at 15:28
SergeSerge
4,13721016
4,13721016
add a comment |
add a comment |
i am not able to understand why we use negative number in array index
That's because
you apparently think
is an array operator, but it is not. It is a pointer operator, defined in terms of pointer arithmetic, which, in a general sense, permits subtracting integers from pointers.you seem to have an expectation of some particular kind of behavior arising from evaluating your example code, but it exhibits undefined behavior on account of performing arithmetic on pointer
pthat does not produce a result pointing into (or just past the end of) the same object thatppoints [in]to. "Undefined" means exactly what it says. Although an obvious program failure or error message might be emitted, you cannot rely on that, or on any other particular behavior. Of the entire program.
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
2
Most noticeably one can also write-1[x]reversing the 'array' and 'index' because the standard saysa[b]is just*(a+b)and + is commutative.
– Goswin von Brederlow
Nov 23 '18 at 16:13
add a comment |
i am not able to understand why we use negative number in array index
That's because
you apparently think
is an array operator, but it is not. It is a pointer operator, defined in terms of pointer arithmetic, which, in a general sense, permits subtracting integers from pointers.you seem to have an expectation of some particular kind of behavior arising from evaluating your example code, but it exhibits undefined behavior on account of performing arithmetic on pointer
pthat does not produce a result pointing into (or just past the end of) the same object thatppoints [in]to. "Undefined" means exactly what it says. Although an obvious program failure or error message might be emitted, you cannot rely on that, or on any other particular behavior. Of the entire program.
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
2
Most noticeably one can also write-1[x]reversing the 'array' and 'index' because the standard saysa[b]is just*(a+b)and + is commutative.
– Goswin von Brederlow
Nov 23 '18 at 16:13
add a comment |
i am not able to understand why we use negative number in array index
That's because
you apparently think
is an array operator, but it is not. It is a pointer operator, defined in terms of pointer arithmetic, which, in a general sense, permits subtracting integers from pointers.you seem to have an expectation of some particular kind of behavior arising from evaluating your example code, but it exhibits undefined behavior on account of performing arithmetic on pointer
pthat does not produce a result pointing into (or just past the end of) the same object thatppoints [in]to. "Undefined" means exactly what it says. Although an obvious program failure or error message might be emitted, you cannot rely on that, or on any other particular behavior. Of the entire program.
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
i am not able to understand why we use negative number in array index
That's because
you apparently think
is an array operator, but it is not. It is a pointer operator, defined in terms of pointer arithmetic, which, in a general sense, permits subtracting integers from pointers.you seem to have an expectation of some particular kind of behavior arising from evaluating your example code, but it exhibits undefined behavior on account of performing arithmetic on pointer
pthat does not produce a result pointing into (or just past the end of) the same object thatppoints [in]to. "Undefined" means exactly what it says. Although an obvious program failure or error message might be emitted, you cannot rely on that, or on any other particular behavior. Of the entire program.
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
answered Nov 23 '18 at 15:32
John BollingerJohn Bollinger
85.1k74279
85.1k74279
2
Most noticeably one can also write-1[x]reversing the 'array' and 'index' because the standard saysa[b]is just*(a+b)and + is commutative.
– Goswin von Brederlow
Nov 23 '18 at 16:13
add a comment |
2
Most noticeably one can also write-1[x]reversing the 'array' and 'index' because the standard saysa[b]is just*(a+b)and + is commutative.
– Goswin von Brederlow
Nov 23 '18 at 16:13
2
2
Most noticeably one can also write
-1[x] reversing the 'array' and 'index' because the standard says a[b] is just *(a+b) and + is commutative.– Goswin von Brederlow
Nov 23 '18 at 16:13
Most noticeably one can also write
-1[x] reversing the 'array' and 'index' because the standard says a[b] is just *(a+b) and + is commutative.– Goswin von Brederlow
Nov 23 '18 at 16:13
add a comment |
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5
You can't, really. It's undefined behavior, so there's no telling what happens. It might work as intended, but it might just as well crash the program or do something even worse. In practice what most likely happens is that it calculates the memory address and just writes/reads to/from it anyway, which may cause data corrupion (because it's overwriting something else), stack corruption (by writing over the stack pointer), or make the OS detect an illegal memory access and crash the program.
– Blaze
Nov 23 '18 at 15:15
Also, you can't assign an integer value to a pointer without a cast.
– EOF
Nov 23 '18 at 15:17