Discussion
What is the output of this C code?
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 incrementing an int * advances the address by sizeof(int), four bytes on a typical implementation. The second address is therefore four higher than the first, which the option with the last digit changing from 8 to c reflects. The particular addresses vary from run to run; the difference of four does not.
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