Why does sizeof(!0) print 1 instead of 4?

Question

#include <iostream>

int main()
{
    std::cout<<sizeof(0);
    return 0;
}

Here, sizeof(0) is 4 in C++ because 0 is an integer rvalue.

But, If I write like this:

std::cout<<sizeof(!0);

here, sizeof(!0) is 1. But, !0 means it print 1, which is also, int type.

then, Why does sizeof(!0) print 1 instead of 4? What am I miss here?

@SidS it means, it is implicitly converted int to bool type. — msc, Oct 06 '18 at 06:35
@M.SChaudhari yes, it receives already the result of the evaluation — Grzegorz Piwowarek, Oct 06 '18 at 06:37
@M.SChaudhari - `sizeof` does not evaluate the expression. It works out the size of the type that would hold the result of the expression. — Peter, Oct 06 '18 at 06:40
Also note that `sizeof(int)` does not have to be `4`, nor does `sizeof(bool) = 1`. But it is enough for identification of this question. — L. F., Oct 06 '18 at 07:36

score 17 · Accepted Answer · edited Jun 20 '20 at 09:12

17

The logical negation operator:

! rhs

If the operand is not bool, it is converted to bool using contextual conversion to bool: it is only well-formed if the declaration bool t(arg) is well-formed, for some invented temporary t.

The result is a bool prvalue.

And sizeof (bool) which is implementation defined is 1 in your implementation.

edited Jun 20 '20 at 09:12

Community

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answered Oct 06 '18 at 06:35

P.W

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Note you are missing a key point which is we are using [sizeof w/ an expression not a type](https://stackoverflow.com/a/18898832/1708801) – Shafik Yaghmour Oct 06 '18 at 14:48
1

`bool operator!(int);` So bool has a size of 1. – Sandburg Oct 10 '18 at 08:50

score 7 · Answer 2 · answered Oct 06 '18 at 06:35

7

!0 is a bool.

sizeof(bool) depends on implementation.

answered Oct 06 '18 at 06:35

Sid S

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Why anybody say `!` is the short for `operator!()` is it that implicit? – Sandburg Oct 10 '18 at 10:00

Grzegorz Piwowarek · Answer 3 · 2018-10-06T07:58:26.520

3

By prepending the integer value with !, you convert it to a boolean - which can be represented using a single byte.

edited Oct 06 '18 at 07:58

answered Oct 06 '18 at 06:36

Grzegorz Piwowarek

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you most probably mean a single *byte*. If you want to get a single bit per boolean, you would need some kind of bitfield. – divinas Oct 06 '18 at 07:30
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@Grzegorz Piwowarek So you expect it to be $\frac{1}{8}$? – L. F. Oct 06 '18 at 07:37
Byte :) also, int is more than 4 bits! – Grzegorz Piwowarek Oct 06 '18 at 07:59

score 1 · Answer 4 · edited Oct 06 '18 at 08:55

1

When we write 0 it is an integer, but when we write !0, it implicitly convert the integer to boolean. ! operator turns any integer into boolean, you can try by writing !1, !2....these all give size of 1 byte. if you wanna know the size of !0 as an int, you can typecast it as

sizeof(int(!0));

edited Oct 06 '18 at 08:55

msc

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answered Oct 06 '18 at 08:22

Akash

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Why does sizeof(!0) print 1 instead of 4?

4 Answers4