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Comments on Behavior of Pointer Arithmetic on the Stack
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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?
Post
I'm able to access a, since b is below a on the stack.
No, it's not!
You have no guarantee in what order the compiler allocates temporary variables on the stack, and even whether it does so at all. You don't even have a guarantee which way (towards high or low addresses) the stack grows. Different compilers on the same machine might do it differently. I've seen this on a PIC 18, for example.
On machines with a lot of registers, both variables might be kept solely in registers when there is little other demand for those registers.
The worst scenario is that whatever is one address past "b" isn't a general memory location, and reading it has side-effects. That's unlikely in this simplified example, but nothing in the standard rules it out. For example, if "b" happened to be allocated to the last register, and that register was mapped into data memory, then something completely unexpected could be at the next address.
Consider an architecture like a Microchip dsPIC. This machine has 16 16-bit registers that are also mapped to addresses 00h to 1Fh in data memory. If the compiler happened to allocate "b" to W14, then the next word is W15, which is the stack pointer. Reading it wouldn't have any side effects in this case, but writing it surely would. On a different architecture, you might end up reading the UART input data register, thereby clearing the last received data. That's unlikely, but you don't know it's not the case without specific knowledge of the machine and the compiler.
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