Question

Why does this code print 4 as the output?

Please also provide some details to help me to understand this type of behaviour better.

int main(){ 
     int *p=NULL;
     printf("%d" ,p+1);
     return 0; 
}

Answer 1

You are getting undefined behavior. It is only valid to add an integer to a pointer if the pointer points to an element of an array or one beyond the end of an array and result of moving the pointer by the number of positions denoted by the integer also points to an element of the same array, or one beyond the end. (A non-array object can be treated as the only element of a one element array for these purposes.)

A null pointer doesn’t point at an element of an array so you can’t add one to it. (In C++ it is explicitly allowed to add 0 to a null pointer value and get a null pointer value as the result. (See this blog entry by Andrew Koenig at Dr. Dobb’s.)

You get further undefined behavior by passing a pointer value to printf where the corresponding format specifier is %d which expects and int. %p is the correct format specifier for void*, you need an explicit cast if you want to print the pointer as an int with %d.

If your program has undefined behavior then there are no guarantees about anything that might happen. You certainly can’t infer anything about the language from observing the results and it is arguable what the merit there is trying to infer properties of the implementation from such behaviour as, without further evidence in the form of observable results from programs which don’t have undefined behavior or guarantees from the compiler vendor, you won’t know what circumstances might cause the observed behavior to change.