Discussion
What is the output of this C code?
#include <stdio.h>
void main()
{
int x = 0;
int *ptr = &x;
printf("%p\n", ptr);
ptr++;
printf("%p\n ", ptr);
}
Answer: A
Correct answer: 0xbfd605e8 0xbfd605ec
Pointer arithmetic is in units of the pointed-to type, so ptr++ advances by sizeof(int) - four bytes on a typical platform. The address rises from ...e8 to ...ec.
The distractor ending e9 assumes a step of one byte, which would be right only for a char *. The particular addresses are arbitrary and vary between runs; what the question tests is the size of the step, which follows from the type.
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