CODEWITHSUNDEEP
c
duslabo
duslabo

Reputation: 1517

Does relational operator affect assignment operator operations?

Why the output of below mentioned program is 0 not 20 ? 

#include <stdio.h>

int main()
{
    int i = 10, j = 0;
    if (i || (j = i + 10))
       /* do something */;                
    printf("%d\n",j);
}

Upvotes: 16

Views: 623

Answers (4)

Grijesh Chauhan
Grijesh Chauhan

Reputation: 58271

Yes, the concept is called Short-Circuit (in logical &&, || operators expression).

In the case of any logical expression (includes ||, &&) compiler stop evaluation expression as soon as result evaluated (and save executions).

The technique for short-circuit is:

!0 || any_expression == 1, so any_expression not need to evaluate.

And because in your expression i is not zero but its 10, so you can think if consdition (i || (j = i + 10)) just as i.

Logical OR 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.

Similarly for && (and operator):
0 && any_expression == 0, so any_expression not need to evaluate.

In your expression: 

(i || (j = i + 10) )
      ------------
       ^
       | Could evaluate if i is 0, 
       as i = 10 (!0 = true), so j remains unchanged as second operand is not evaluated

For or || operator answer can be either 0, 1. To save execution, evaluation stops as soon as results find. So if first operand is non-zero result will be 1 (as above) for the expression. So for first operand i = 10 compares unequal to 0, the second operand (j = i + 10) is not evaluated so j remains 0 hence output of your code is 0.

Note: Short-circuit behavior is not only in present in C but concept is common to many languages like Java, C++, Python. (but not all e.g. VB6).

In C short-circuiting of logical expressions is guaranteed has always been a feature of C. It was true when Dennis Ritchie designed and implemented the first version of C, still true in the 1989 C standard, and remains true in the C99 standard.

A related post: Is short-circuiting boolean operators mandated in C/C++? And evaluation order?

Upvotes: 27

tczf
tczf

Reputation: 85

because || is a short-circuit operator (so is the && operator).

so in (i || j = i+10), i is 10, left part of || is true, the expression j = i+10 didn't happen, as a result, j=0.

Upvotes: 3

johan d
johan d

Reputation: 2863

In if (i || (j = i + 10)), there are two boolean expression to evaluate. The thing is, the first one is true, therefore there is no need to compute the second. It's purely ignored.

Upvotes: 5

Paul R
Paul R

Reputation: 212979

|| is a short-circuit operator - if the left hand side evaluates to true then the right hand side does not need to be evaluated. So, in your case, since i is true then the expression j = i + 10 is not evaluated. If you set i to 0 however then the right hand side will be evaluated,

Upvotes: 20

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