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Consider the following code: #include <stdio.h> int main() { int a = 5; int b; ++*(&b + 1); printf("%d\n", a); return 0; } The output is as expected: 6 ...
#3: Nominated for promotion
by
Alexei
·
2022-02-13T11:38:57Z (about 2 years ago)
#2: Post edited
by
Alexei
·
2021-12-14T06:23:55Z (over 2 years ago)
replaced tag with a more specific one
#1: Initial revision
by
Josh Hyatt
·
2021-12-13T22:33:58Z (over 2 years ago)
Behavior of Pointer Arithmetic on the Stack
Consider the following code:
```
#include <stdio.h>
int main() {
int a = 5;
int b;
++*(&b + 1);
printf("%d\n", a);
return 0;
}
```
The output is as expected:
```
6
```
By creating and incrementing a pointer to `b`, I'm able to access `a`, since `b` is below `a` on the stack. Is this behavior guaranteed by the C language, or is this undefined/unspecified behavior? If UB, what does the standard have to say that disallows this? For example, does C guarantee that the stack grows downwards, or that arithmetic with pointers into the stack is valid?