C++ Tilde Operator on bool

Question

I've done the LinkedIn C++ Assessment and got the following question:

What is the result from executing this code snippet?

bool x=true, x=false;

if(~x || y) {
    /*part A*/
}
else {
   /*part B*/
}

I don't know anymore what the answers were, but I thought "B" should be displayed, right?

I thought by "~", x is inverted ("bitwise NOT") and is therefore no longer "true"

But when I run the code, I get "A":

Code:

#include <iostream>

int main()
{

    bool x = true, y = false;

    std::cout << "x: " << x << std::endl;
    std::cout << "~x: " << ~x << std::endl;

    if (~x || y) {
        std::cout << "A" << std::endl;
    }
    else {
        std::cout << "B" << std::endl;
    }

    return 0;
}

Output:

x: 1
~x: -2
A

Can someone explain this to me?

`false` is `0` and `true` is anything else, in your case the compiler initialized x to `1`. The bitwise complement of `1` is not `0` but a value with all bits set except the last one (which as an int would be -2). Thus you still get `true` — perivesta, Sep 06 '21 at 21:28
@dave: No, `true` is not "anything else". Conversion of `bool` to any integral type will always produce exactly `+1` for `true`. The conversion happens here due to applying "the usual promotions" to operands. — Ben Voigt, Sep 06 '21 at 21:29

score 2 · Accepted Answer · edited Sep 06 '21 at 21:47

In this expression

~x

there is applied the integral promotions to the operand x of the type bool. The result of the promotion is an object of the type int that has the value equal to 1 like (in binary)

00000000 00000000 00000000 00000001

The operator ~ inverses bits and you will get

11111111 11111111 11111111 11111110

From the C++ 14 Standard (5.3.1 Unary operators)

10 The operand of ~ shall have integral or unscoped enumeration type; the result is the one’s complement of its operand. Integral promotions are performed. The type of the result is the type of the promoted operand.

and (4.5 Integral promotions)

6 A prvalue of type bool can be converted to a prvalue of type int, with false becoming zero and true becoming one.

7 These conversions are called integral promotions.

As this value is not equal to 0 then used as an operand of the logical OR operator it is implicitly converted to the boolean value true.

From the C Standard (5.15 Logical OR operator)

1 The || operator groups left-to-right. The operands are both contextually converted to bool (Clause 4). It returns true if either of its operands is true, and false otherwise. Unlike |, || guarantees left-to-right evaluation; moreover, the second operand is not evaluated if the first operand evaluates to true.

and (4.12 Boolean conversions)

1 A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true. For direct-initialization (8.5), a prvalue of type std::nullptr_t can be converted to a prvalue of type bool; the resulting value is false.

C++ Tilde Operator on bool

1 Answers1