[SOLVED] use of unsigned cast with array pointers

atlantis43 · 09-21-2013, 09:41 PM

Hoping that someone can explain why (unsigned) cast is needed in the following code which gives result of 4,while leaving out cast yields result 1:

Code:

#include <stdio.h>

int main(void){

int array2[2];
int *ptr3;

ptr3 = array2;
printf("%d\n", (unsigned)(ptr3+1)-(unsigned)ptr3);
printf("%d\n", (ptr3+1)-ptr3);

return 0;
}

rknichols · 09-21-2013, 10:11 PM

Sure. By definition, the expression "(ptr3+1)-ptr3" yields a value which when added to ptr3 yields ptr3+1, and that value is simply 1 since the scaling by sizeof(*ptr3) is implied in the addition.

When you do the (unsigned) typecast, the pointer type is lost, so the result is a number which when added to an integer containing the bit pattern of ptr3 will yield an integer containing the bit pattern of (ptr3+1). The addition is done on the pointer type, and gets the implied scaling, but the subtraction is simply the difference between two integers.