Why the two following C code differ by output?

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The following code outputs 4:

C++

#include<stdio.h>
int main()
{

    printf("%d",1+(int)NULL);
}

While the following code outputs1:

C++

#include
int main()
{

    printf("%d",1+(int)NULL);
}

What I have tried:

I don't understand why the above codes differ differ?

Posted 12-Mar-16 4:34am

Member 12378355we

Updated 12-Mar-16 5:14am

CHill60

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CHill60 12-Mar-16 10:53am

What was the #include in the second program and are you sure that you have posted both programs here correctly?
Are these running on the same machine / operating system?

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OriginalGriff · Accepted Answer · 2016-03-12T05:14:00

Probably because stdio.h #defines NULL to a specific value if it isn't defined.
The standard VS C++ stdio.h uses this:

C++

/* Define NULL pointer value */
#ifndef NULL
#ifdef __cplusplus
#define NULL    0
#else  /* __cplusplus */
#define NULL    ((void *)0)
#endif  /* __cplusplus */
#endif  /* NULL */

But it would be possible to define a NULL that still ended up as a void pointer after the addition of the (int) cast. And since a pointer + 1 increments by the size of a "machine word", it's quite likely that in a 32 bit environment, the resulting address would be four.
I'd start by looking at the exact content of your include files and check what they define NULL as (since it isn't defined as a C or C++ keyword)