Question

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!

Answer 1

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).