Question

Take the following C code as an example:

    char buffer1[5];
    int* ret;
    printf("Buffer1 is: %x\n", (int*)buffer1);
    ret = (int*)buffer1 + 12;
    printf("Ret is: %x\n", ret);

I just want to add 12 bytes to buffer1 and store it in ret. This is obviously wrong, but I don’t know how to properly add hex address in C. Any help is appreciated.

Answer 1

If you want to add 12 bytes, you need a data type that work on byte boundaries, like a char *. Now you may think you’re doing that (buffer1 decays to a char *) but, because a cast binds more tightly than an addition, (int*)buffer1 + 12 actually means ((int*)buffer1) + 12 rather than (int*)(buffer1 + 12).

And the problem with adding 12 to an int * is that it scales the addition. If your int is four bytes, adding 12 to it will actually add 48 bytes.

If you change the assignment line to:

ret = (int*)(buffer1 + 12);

you will find that the addition happens to buffer1 so will not be scaled, then the cast to an int * will operate on that value.

I should mention of course that dereferencing the resultant pointer is probably not a good idea since it will be beyond the bounds of the actual array.

I’ll also suggest that %p is probably a better format string to use for pointers.