I don't understand how to evaluate w1 and w1. I know i>1 is 0, j<0 is 0, i<0 is 1, j>0 is 1 and so on, but how do I associate these values?

#include <stdio.h>

int main(void)
{
    int i = -1, j = -i;
    int w1, w2;
    w1 = (i > 0) && (j < 0) || (i < 0) && (j > 0);
    w2 = (i <= 0) || (j = 0) && (i >= 0) || (j <= 0);
    printf("%d", w1 == w2);
    return 0;
}

w1=1 and w2=0 but I know that is incorrect. Can someone explain in detail the process?

CodePudding user response：

Check the operator precedence to have a complete view.

In your case you have this order. First (<,>,<=,>=), then && and finally ||.

So:

w1 = false && false || true && true -> false || true -> true (1)
w2 = true || false && false || false -> true || false || true -> true (1)

and w1==w2 = true (1)

CodePudding user response：

Short-circuiting:

The logical AND and OR operators short-circuit.

From C11:

The AND logical operator:

The && operator shall yield 1 if both of its operands compare unequal to 0; otherwise, it yields 0. The result has type int.

4 Unlike the bitwise binary & operator, the && operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares equal to 0, the second operand is not evaluated.

6.5.14 Logical OR operator

3 The || operator shall yield 1 if either of its operands compare unequal to 0; otherwise, it yields 0. The result has type int.

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

Assignment vs comparison:

j = 0

assigns 0 to j. It doesn't test for equality.