Why does double negation force values to become a bool?

Question

If x was a 4 bit word like 1010, and you did the operation !!x,

Wouldn't the first !x return 0101,
And the second !(!x) return 1010?

Rather it returns ...0001 or ...0000. Why is this?

Related: https://stackoverflow.com/questions/9229601/what-is-in-c-code (I swear there was a better question like this but can't find it at the moment..) — scohe001, Feb 07 '18 at 19:57
First `!` turns any non-zero value into `false`, second `!` makes it `true` (and vice versa) — Killzone Kid, Feb 07 '18 at 20:09
In C, "double negation force values to become a bool" is not quite correct. `!` result is an `int`, not a `bool`, with a value of 0 or 1. — chux - Reinstate Monica, Feb 07 '18 at 22:27

Bathsheba · Answer 1 · 2018-02-07T20:01:19.277

7

In C, !x is either 1 or 0, So !!x is a "collapse to 0 or 1 operator" in the sense that any non-zero number is mapped to 1, and 0 stays as it is. This can be useful on occasions.

In C++, !x is a bool type, So !!x is a "collapse to false or true operator" in the sense that any non-zero number is mapped to true, and zero is mapped to false.

edited Feb 07 '18 at 20:01

answered Feb 07 '18 at 19:54

Bathsheba

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@Adrian I scrapped the c#. Perhaps you can answer for that one. – Bathsheba Feb 07 '18 at 20:01
The question does not really apply to c# anyway. C# does not have implicit integer to bool conversion. – adjan Feb 07 '18 at 20:03
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@Adrian Saved by the retagging! – Bathsheba Feb 07 '18 at 20:06
Why would you not simply do `(bool)x` or `(_Bool)x`? – Aykhan Hagverdili May 31 '20 at 23:25
@Ayxan: That changes the type of the expression. – Bathsheba Jun 01 '20 at 06:00
@Bat `!!x` does too in C++. Also, does it ever matter in C? – Aykhan Hagverdili Jun 01 '20 at 09:21

score 6 · Answer 2 · answered Feb 07 '18 at 20:02

The ! operator performs logical negation. If its argument is non-zero, it results in 0. If its argument is 0, it results in 1.

What you're describing is the bitwise complement operator, denoted by ~.

These are both described in section 6.5.3.3 of the C standard:

4 The result of the ~ operator is the bitwise complement of its (promoted) operand (that is, each bit in the result is set if and only if the corresponding bit in the converted operand is not set). The integer promotions are performed on the operand, and the result has the promoted type. If the promoted type is an unsigned type, the expression ~E is equivalent to the maximum value representable in that type minus E.

5 The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. The result has type int. The expression !E is equivalent to (0==E).

I think C's `!` can be better described as a "!=0" operator. When applied to logical values it negates them, but it may also be applied to other types such as floating-point values or pointers. — supercat, Feb 07 '18 at 20:05

kwyntes · Answer 3 · 2018-02-07T20:08:21.537

2

! is a boolean operator, so it will convert the number into a bool (any non-zero value is true, zero is always false). That other ! inverts the bool.

edited Feb 07 '18 at 20:08

answered Feb 07 '18 at 19:56

kwyntes

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In C (as post is dual tagged - sigh), `!` result is not a `bool`, but an `int`. – chux - Reinstate Monica Feb 07 '18 at 22:25
Yes, in C there are no `bool`s (not like `true`/`false`) but it's an `int` value. (either `0` or `1`) but in C++ and C# it is a `bool` – kwyntes Feb 08 '18 at 15:42
Disagree with "in C there are no bools (not like true/false)". Since C99, C has a boolean type `_Bool` (aka `bool` via ``). The C return type of `!` still remains `int`. – chux - Reinstate Monica Feb 08 '18 at 15:48
Sorry, didn't know that. – kwyntes Feb 08 '18 at 15:54
Posts tagged C _and_ C++ unfortunately create issues like this - not your fault. – chux - Reinstate Monica Feb 08 '18 at 15:56

Why does double negation force values to become a bool?

3 Answers3