Preincrement and Logical Operator

**NKP** · 01-25-2010

Hi,

I have the following code snippet:

Code:

#include<stdio.h>
int main()
{
int x,y,z;
x=y=z=1;
z= ++x||++y&&++z;
printf("x=%d y=%d z=%d\n", x,y,z);
return 0;
}

And the output is :
x=2 y=1 z=1

I cannot understand why y is 1, even if it is pre incremented?
Can any one help?

**Adak** · 01-25-2010

Code:

z= ++x||++y&&++z;

In the red part of this line of code, you're assigning an incremented x to z. So that part of the line evaluates to TRUE or 1.

So the blue part, is never executed.

**laserlight** · 01-26-2010

By the way, if x has the value of -1 and y does not have the value of -1, that statement actually results in undefined behaviour since you would be incrementing z and assigning the result to z, with no sequence point in between.

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