CODEWITHSUNDEEP
clogical-operators
Anup
Anup

Reputation: 11

why does the value of num2 in below expression becomes -1 instead of 0?

The below C program gives output -1 for num2. I do not understand why. Can somebody help me with it?

#include <stdio.h>

int main(void)
{
    int num1 = 0, num2 = -1, num3 = -2, num4 = 1, ans;
    ans = num1++ && num2++ || ++num4 && num3++;
    printf("%d %d %d %d %d", num1, num2, num3, num4, ans);
    return 0;
}

output:

1 -1 -1 2 1

Upvotes: 0

Views: 107

Answers (2)

chqrlie
chqrlie

Reputation: 144951

The expression num1++ && num2++ || ++num4 && num3++ is parsed as:

(num1++ && num2++) || (++num4 && num3++)

Both || and && use shortcut evaluation, meaning that the right operand if and only if the left operand evaluates to false (resp. true).

  • num1++ && num2++ is evaluated first:
  • num1++ is evaluated: it evaluates to the initial value of num1, 0 and num1 is incremented.
  • since the left operand if false, the right operand of && is not evaluated (num2 is unchanged`) and the expression is false:
  • num1++ && num2++ is false, so the right operand of || must be evaluated to determine the value of the whole expression:
  • ++num4 is evaluated: the value is 2 and num4 is incremented.
  • the left operand of this second && operator is true, so the right operand is evaluated
  • num3++ is evaluated: the value is -2 and num3 is incremented.
  • 2 && -2 is true, the whole expression evaluates to true, which has type int and the value 1 in C: ans receives the value 1.
  • printf outputs 1 -1 -1 2 1 (without a trailing newline)

Upvotes: 1

some-user
some-user

Reputation: 4924

If the left side of a logical AND && is false, the right side will not be evaluated as the result can be determined as false already.

Upvotes: 0

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