CODEWITHSUNDEEP
c
Apoorva Somani
Apoorva Somani

Reputation: 515

C Program output confusion

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;
}

Upvotes: 6

Views: 222

Answers (4)

Alejandro Caro
Alejandro Caro

Reputation: 1092

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

Upvotes: 0

Vikram S. Parikh
Vikram S. Parikh

Reputation: 11

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!!!

Upvotes: 0

Spikatrix
Spikatrix

Reputation: 20244

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.

Upvotes: 1

Rizier123
Rizier123

Reputation: 59701

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

Upvotes: 6

Related Questions