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cpointerspointer-arithmetic

Pointer Arithmetic In C

Consider the following code fragment:

int (*p)[3];
int (*q)[3];

q = p;
q++;
printf("%d, %d\n", q, p);
printf("%d\n", q-p);

I know that pointer arithmetic is intelligent, meaning that the operation q++ advances q enough bytes ahead to point to a next 3-integers-array, so it does not surprises me that the first print is '12, 0' which means that incrementing q made it larger in 12.

But the second print does surprises me. It prints 1!
So why would it print 1 instead of 12? it just puzzles me.

Solution

  • Like the ++ increment operator, the - subtraction operator with pointers also takes into account the size of the objects being pointed to. Specifically, the result returned is the number of bytes difference in the pointer values divided by the size of the pointed-to object (12, in your example). So the difference is 12 bytes, divided by size 12, or 1.