Different Output

[cyrus]

My C program is as below:
#include <stdio.h>
main()
{
int x=900,y=400,z=0;
if(!x>=1000)
y=500;
z=100;
printf("y=%d z=%d",y,z);
}

I thought the output would be
y=500 z=0
But to my surprise the output was
y=400 z=100

Why is it so?

[Mukhtar Ahmad]

#include <stdio.h>
main()
{
int x=900,y=400,z=0;
if(!x>=1000)
y=500;
z=100;
printf("y=%d z=%d",y,z);
}

I thought the output would be
y=500 z=0
But to my surprise the output was
y=400 z=100


It the place where operator precedence and associativity come to play its part. in the if condition ((!x>=1000) !x is evaluated first then ( result of !x>= 1000) eventually resulted as false .
as condition is applied to only one statement i.e. y=500; which will not be processed in the very case.
You have the result as it displayed.

[zsk_00]

#include <stdio.h>
main()
{
int x=900,y=400,z=0;
if(!x>=1000) // condition is true
y=500; // ok as this is part of if statement and contion is true
z=100; // this will always be executed as its not part of if statement.
printf("y=%d z=%d",y,z);
}

so
y=500 and z=100 is what you should get.

The following will give what you were expecting.

#include <stdio.h>
main()
{
int x=900,y=400,z=0;
if(x>=1000) { // note that ! have been removed
y=500;
z=100;
}
printf("y=%d z=%d",y,z);
}

[caradoc]

The evaluation of the statement
if(!x>=1000)
takes place as follows:
Since the precedence of ! is greater than >= it gets evaluated first and so !x is evaluated first. Since x=900 !x results in 0 and 0>=1000 becomes false and so statement jumps to z=500 and your y value remains the same as 400.