C Program output confusion

Question

Can someone explain why the output of this program is false??

x && y gives 1. Still the output is false.

#include <stdio.h>

int main()
{
    int x = 1, y = 2;
    if(x && y == 1)
    {
        printf("true.");
    }
    else
    {
        printf("false.");
    }

    return 0;
}

Because of operator precedence rules, x && y == 1 = x && (y == 1) — David Ranieri
– David Ranieri, Commented Feb 24, 2015 at 16:07
You probably meant something like (x && y) == 1.. but x && y is preferable. — Daniel Kleinstein
– Daniel Kleinstein, Commented Feb 24, 2015 at 16:07

Rizier123 · Accepted Answer · 2015-02-24 16:06:34Z

6

Because == has a higher precedence than && So first this get's evaluated:

x && (y == 1)

y == 1  // 2 == 1
//Result: false

Which is false and then second:

x && false  //1 && false
//Result: false

So the if statement will be false

For more information about operator precedence see here: http://en.cppreference.com/w/cpp/language/operator_precedence

answered Feb 24, 2015 at 16:06

Rizier123

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Comments

Vlad from Moscow · Accepted Answer · 2015-02-24 16:09:51Z

1

if(x && y == 1)

Is the same as

if( ( x != 0 ) && ( y == 1 ) )

Here,x != 0 is true, but y == 1 is false. And since at least one of the operands of && is false, the condition evaluates to false and the else part executes.

edited Feb 24, 2015 at 16:09

Vlad from Moscow

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answered Feb 24, 2015 at 16:06

Spikatrix

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2 Comments

juanchopanza Over a year ago

It doesn't really help if you don't say anything about operator precedence.

Spikatrix Over a year ago

Yeah . You're right. I mis-interpreted the question a bit and did not see "x && y gives 1. Still the output is false.". Now @Rizier123 has a correct answer to the question.

Vikram S. Parikh · Accepted Answer · 2015-02-25 07:24:51Z

0

It clearly stated X = 1 & Y = 2; Now with your expression

X && Y == 1

The expression is evaluated as Y == 1 (Precedence Rule, Also output is False)

X != 0 (Its True)

Now && is Logical And Operator, so it evaluates to True only if both the parts in expression evaluates to True!!!

answered Feb 25, 2015 at 7:24

Vikram S. Parikh

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Comments

Alejandro Caro · Accepted Answer · 2016-07-03 23:22:11Z

0

It's okay to false, then 2 and 2 and it is different from one. What you're asking is whether both x and y both are worth 1. If this happens say true but false

answered Jul 3, 2016 at 23:22

Alejandro Caro

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