CODEWITHSUNDEEP
c++
Ritesh Kumar Gupta
Ritesh Kumar Gupta

Reputation: 5191

Precedence Rule

I was solving the C++ Multiple choice questions. I am not able to understand the output for the following code::

#include <iostream>
using namespace std;
int main()
{
    int x,y,z;
    x=y=z=1;
    z=++x || ++y && ++z;
    cout<<x<<" "<<y<<" "<<z<<endl;
    system("pause");
    return 0;
}

I am solving this question in following way:: Precedence order :: 

Precedence "++" greaterthan Precedence "&&" greaterthan  Precedence "||"

Also , the Associativity of unary++ is "Right to left" . So

z=(++x)||(++y) && (2)
z=(++x)||(2)&& (2)
z=(2)||(2)&&(2)
z=(2)|| 1  //As 2 && 2 is 1(true)
z=1       // As 2 || 1 is 1(true)

So as per me ,the correct output should be x=2,y=2 and z=1.

But When i ran this code in my compiler,the compiler output is x=2,y=1,z=1.

Why i am getting such output and where i am making mistake?

Thanks!

Upvotes: 0

Views: 317

Answers (4)

Nathan822
Nathan822

Reputation: 198

Keep in mind that the compiler is going to execute the whole logical expression only if it NEEDS to. Otherwise, it tries to get away with as little work as possible.

For example, if you use the && operator, it is necessary for both expressions to be evaluated. However, if you use the || operator (which you have), if the first expression on the left hand side is true, the right hand side is not executed. In your example, ++x gives 2, which is evaluated to a boolean TRUE. End of story for the compiler since once true, an OR statement will never revert to FALSE. This concept is called 'short-circuiting' by the compiler. That is why you got the output that you described.

Upvotes: 1

Jainendra
Jainendra

Reputation: 25153

Logical OR Operator requires if any of the two operands is non zero then then condition becomes true. In the expression A||B, either A or B is non zero then (A || B) will br true or equal to 1. So ++y and ++z will be ignored by the compiler because value of ++x is 1.

Upvotes: 1

CB Bailey
CB Bailey

Reputation: 792867

Operator precedence tells you how to group expressions; it doesn't tell you in which order they are executed.

|| and && are special in that the first operand is always evaluated first and the second operand (including all sub-expressions) is only evaluated if it is required to determine the value of the expression.

For ||, if the first operand evaluates to true the second operand is not evaluated because the result of the logical-or will always be true.

Similarly, the second operand of && will not be evaluated if the first operand evaluates to false as the logical-and must be false in this case.

In the expression z=++x || ++y && ++z, the grammar rules specify a grouping:

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

In the sub-expression (++x) || ((++y) && (++z)), as (++x) evaluates to true (as 2 is non-zero), the second operator ((++y) && (++z)) is never evaluted. x becomes 2, y is unchanged and z is assigned 1 (true converted to an integer).

Upvotes: 3

Kevin Reid
Kevin Reid

Reputation: 43882

The && and || operators are short-circuiting. They evaluate the left side, and then evaluate the right side only if necessary to determine the value.

Therefore, if the left side of || is true the right side is not evaluated, and if the left side of && is false the right side is not evaluated.

In your example, since ++x is true (2), the right side of the || is not evaluated.

Upvotes: 3

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