Question

Plain C, on Windows 7 & HP machine.

int main(void) {

    unsigned int a =  4294967295;
    unsigned int *b = &a;

    printf("before val: '%u'\n", *b); // expect 4294967295, got 4294967295

    memset(b+2, 0, 1);

    printf("after val: '%u'\n", *b);
    // little endian          4th      3rd     2nd       1st
    // expect 4278255615 - 11111111 00000000 11111111 11111111
    // got    4294967295 - 11111111 11111111 11111111 11111111

    return 0;

}

I want to set the third byte of the integer to 0x0, but is remains the same. Any ideas? Thank you.

On my machine, int is 32 bits.

Answer 1

Pointer addition/subtraction does not move by only one byte – it moves by the size of the type of the object being pointed to.

That is to say (assuming 4-byte integers),

int *p = 0x00004
int *q = p+1;
assert(q == 0x00008)

Basically, it’s the same as if you used the index of operator:

int *q = &p[1]

If you want to increment a pointer by one, cast it to a unsigned char *. The way you did it, you were overwriting memory that was not part of the variable a and possibly overwriting existing data for something else.