If bitwise OR operator is used twice between two number. How will that work?

Question

#include <stdio.h>
int main()
{
    int a = 60; // 0011 1100
    int b = 13; // 0000 1101
    int c = 0;
    c = a || b;
    printf("%d",c);
    return 0;
}

The output for my code is 1. Can anyone explain how that worked?

Answer 1

In this statement

 c = a || b; // 0011 1101

there is used the logical OR operator || that yields 1 if either of operands is unequal to 0.

From the C Standard (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.

The bitwise inclusive OR operator | is written like

 c = a | b; // 0011 1101

If you will write like

 c = a | | b;

with a blank between the symbols '|' then the compiler will issue an error. You may not use such a way any binary operator because such an operator expects operands on the left and the right sides of the operator.

Of course, if you will write for example like

c = a + + b;

then there is the first operator + is the binary plus and the second operator + is the unary plus operator. That is there is no two consecutive binary operators +.

You should not mix logical operators || and && with bitwise operators | and &.