Logical OR operand evaluation basics in C

Question

I'm learning C basics and I'm trying to understand why the output of this example is 1 2 1 1:

int main()
{
    int a, b, c, d;
    a = b = c = d = 1;
    a = ++b > 1 || ++c > 1 && ++d > 1;
    printf("%d%d%d%d", a, b, c, d);
}

As far as I understand, only the left operand before Logical OR || is getting evaluated because it is True and there is no need to evaluate the right one. The right operand is not evaluated. Is this right?
The left operand is true because ++b > 1 equals 1 (2 is larger than 1) and a = 1. Right?

For future reference, the term you are looking for is "short circuiting". — StoryTeller - Unslander Monica, Sep 04 '16 at 12:40
The left side of the `||` evaluates to true, and therefore the right side does not need to be evaluated. Hence, `c` and `d` remain `1`, `b` is `2`, and `a` gets assigned to true, which is `1`. — Tim Biegeleisen, Sep 04 '16 at 12:42

score 2 · Answer 1 · edited May 23 '17 at 12:22

2

Yes, you are right on both accounts.

The important to note is that this behaviour is guaranteed by the C standard and is commonly known as short circuting.

Relevant: Is short-circuiting logical operators mandated? And evaluation order?

edited May 23 '17 at 12:22

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answered Sep 04 '16 at 12:41

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score 1 · Accepted Answer · edited May 23 '17 at 12:32

The left side of the || evaluates to true, and therefore the right side does not need to be evaluated. Hence, c and d remain 1, b is incremented to 2, and a gets assigned to true, which is promoted to 1.

Explanation:

a = b = c = d = 1;

All four variables equal 1.

a = ++b > 1

The variable b gets incremented to 2 before the rest of the RHS gets evaluated. This results in the expression 2 > 1, which is true. As others have mentioned, your || expression is then short-circuited, which means that what follows the || will not be evaluated because the program already knows that the entire RHS will be true regardless of what happens. This is why I did not bother to even write the expression ++c > 1 && ++d > 1, because it will be skipped.

And the variable a is set to true, which is the same as 1.

Reference on promoting true to int: Can I assume (bool)true == (int)1 for any C++ compiler?

score 1 · Answer 3 · answered Mar 17 '18 at 13:35

The first line a = b = c = d = 1 assigns 1 to all variables as the associativity of the assignment operator (=) is from right to left.

Next: The concept to be applied here is called short-circuiting. The preference of '>' is higher than all the other operators except for the increment operator(++) -- the prefix increment to be precise.

Hence, first b gets incremented to 2. Now, the comparison of 2 > 1 occurs, which is true or 1. The left operand of the OR is true or 1. Thus, it hardly makes a difference if the right operand is true or false as the final output would still return true or 1.

Thus, everything to the right of the OR (||) isn't even evaluated. So c and d are not incremented and remain = 1.

a = True or 1 Thus a = 1, b = 2, c = d = 1

score 0 · Answer 4 · answered Jan 02 '19 at 01:15

OR operator in c is the Logical operator. || Logical OR. True only if either one operand is true. In first line you assign 1 to all variables by a=b=c=d=1. After that, you are comparing conditions that a=++b>1 if it is true then the compiler will check the second condition that ++c>1 && ++d>1 this is with the && AND operator so in this if both equations are correct only then the command will run.

Logical OR operand evaluation basics in C

4 Answers4