Discussion
What is the output of this C code?
#include <stdio.h>
void main()
{
int x = 0;
int *ptr = &x;
printf("%d\n", *ptr);
}
Answer: D
Correct answer: 0
ptr holds the address of x, and *ptr reads the object there, which is 0 - the value x was initialised with.
The %d conversion prints it as an integer. Printing the pointer itself would need %p and would show an address, which is the first distractor. There is nothing uninitialised here, so no junk value arises, and dereferencing a pointer to a live object is entirely well defined.
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