What does 'p2=p1+(!!p3)' mean? 1

[pjw001]

The fields are defined as

char *p1,*p2,p3;

What does the statement

p2=p1+(!!p3);

do?

The program does work but I cannot find any documentation that defines the meaning of '!!'.

Thanks in anticipation.

 

[amarg]

Hi,

p2=p1+(!!p3);

In this statement you are incrementing the address contain in p1 by 1 or 0 based on the initial value in p3 and finally storing in p2.

For example p3 contains other that '\0' it means !p3 is non zero and !!p3 is again equal to 1, so statement will store the address contain in p1 plus 1 in p2.

Now other example p3 contains '\0' it means !p3 is equal to 1 and !!p3 is equal to 0, so statement will do nothing but copy the p1 in p2.

Just let me know if you have any doubt.

 

[pjw001]

Thank you very much.

I had actually considered the concept of the ! signs effectively reversing their effect in some way, but decided that this was too silly to persue!

Bring back cobol, all is forgiven.

 

[amarg]

This example will help you to get the meaning.

#include <stdio.h>

void main()
{
char *p1 = &quot;AMAR DEEP GAUR&quot;;
char *p2 = NULL;
char p3;

p3 = '\0';
p2=p1+(!!p3);

printf(&quot;when p3 is null p2 is pointing to the character '%c'\n&quot;, *p2);

p3 = 'A';
p2=p1+(!!p3);

printf(&quot;when p3 is not null p2 is pointing to the chargacter '%c'\n&quot;, *p2);
}


Output is

when p3 is null p2 is pointing to the character 'A'
when p3 is not null p2 is pointing to the chargacter 'M'