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ccomparison-operatorspost-increment

What is the order of evaluation of the statements in this code?

I came across this question.

#include <stdio.h>
int main()
{
int k=8;
int x=0==1||k++;
printf("%d  %d",x,k);
    return 0;
}

The output is 1 9.

As this answer suggests that

Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated.

I am unable to understand how the statement int x=0==1||k++ is evaluated,due to which the value of x and k becomes 1,9 respectively.

Can someone explain how such statements are evaluated by the compiler in c ?

Solution

  • "Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated."

    yes it's true ...

    to make it clear first make sure that you are aware of these basics

    1) (1 || any_var) is 1

    2) Operator precedence is as follows

    ++

    then

    ==

    then

    ||

    NOW coming to your doubt of || vs |

    note that | (single pipe) operator ..,will execute both LHS and RHS , no matter what)

    whereas

    || (double pipe) evaluates LHS of || first and if it is 1 it need not evaluate RHS (for speed)

    RHS of || operator will not be evaluted if LHS comes out to be true.

    but here 0==1 is false i.e 0==1 returns 0

    hence RHS will be evalauted

    so the statement

    k++ is executed

    but 8 is used (as property of post increment operator says--> first use then increment)

    so 0||8 is definitely true (1) so x evaluates to 1

    and then k is incremented after sequence point ie k is made equal to 9

    hence output x=1 and k=9

    I hope it clears your doubt :)