Question

Possible Duplicate:
void * arithmetic

Hi guys I have a small question regarding pointer increment in C. First let say that I know that ptr++, where ptris a pointer will increment as much as the sizeof(*ptr). Also I know that when doing *ptr, the compiler knows it has to grab sizeof(*ptr) bytes from memory.

The second part helps me understand why the following does not even compile:

int main(){
 int a = 3;
 void* b = &a;
 printf("%d\n", *b);
 return 0;
}

Because the compiler does not know the size of a variable of type void. However, I’m a little bit confused about the following code:

int main(){
 int a = 3;
 void* b = &a;
 printf("%p\n", b);
 b++;
 printf("%p\n", b);
}

So, my two questions are:

1. How is the compiler able to know how much it should increment b?

2. Why does it increment only one byte (at least in my machine is one byte)?

Answer 1

1) it doesn’t, 2) that’s undefined behaviour. void is an incomplete type, so it doesn’t have a well-defined size, so you cannot do pointer arithmetic with its pointers.

Typically you will want char pointers for byte-wise memory fiddling.

If you compile with all compiler warnings enabled, you will spot such problematic code.