Thread: Confusion with usage of ! operator in if statment

See example below:

Code:

#include <stdio.h>
#include <setjmp.h>

static jmp_buf buf;

void second(void) {
    printf("second\n");         // prints
    longjmp(buf,1);             // jumps back to where setjmp was called - making setjmp now return 1
}

void first(void) {
    second();
    printf("first\n");          // does not print
}

int main() {   
    if ( ! setjmp(buf) ) {
        first();                // when executed, setjmp returns 0
    } else {                    // when longjmp jumps back, setjmp returns 1
        printf("main\n");       // prints
    } 
    return 0;
}

I don't understand this part "if ( ! setjmp(buf) )" does. Can anyone explain?

Place the code in code tags like this [code]/*YOUR CODE*/ [/code]

Well we normally write what you have there, when the result is zero.

Example.

Code:

int a = 0;
if( a == 0)
{
     //it's zero
}
else
{
     // it's not zero
}

Many people write this code we just read like this (which is equivalent)

Code:

int a = 0;
if( !a )
{
     //it's zero
}
else
{
     // it's not zero
}

a is zero, so we have as condition of if this: !0 which is 1 which results to true.

It's nice to use this style when you have variables names found, etc. ...

But notice that you have a .h file that defines what the function set returns.

Remember: 0 results to false, anything else in true.

You are welcome

This is something that had to do with the implementation of the function. This is something of the standard. You created setjmp.h or is it something that everybody around you use it?