Thread: Confused about an uninitialized bool variable.

Hello.

Sparing you the details of how I have come across this issue I am just going to paste an example which I simply do not get (although it's probably easy to explain - knowing my luck it has something to do with undefined behaviour

Code:

#include <stdio.h>
#include <stdbool.h>

int main(void)
{
    bool b;

    if (b)
        printf("%d: True\n", b);
    else
        printf("%d: False\n", b);

    if (!b)
        printf("%d: False\n", b);
    else
        printf("%d: True\n", b);

    return 0;
}

Running the program results in:

Code:

assiduus@ubuntu:~$ ./a.out
183: True
183: False

When I manually assign this random value to b, 1 is stored instead, which I understand is a normal behaviour when you try to store a nonzero value into b. The result is what you would expect:

Code:

assiduus@ubuntu:~$ ./a.out
1: True
1: True

How come the same thing does not happen during the first (b uninitialized) run? And how is it that 183 is true or false depending on how you form the "question"?

Cheers,

Please see this link: bool

Jim

There is no requirement that _Bool only be capable of storing 0 and 1; it must simply be able to hold both those values.

As you imply, converting a scalar to _Bool requires that the value be converted to 0 or 1, but this is only on conversion to _Bool. So _Bool is probably implemented as a char, with special code to handle assignment/conversion to _Bool. However, _Bool is likely capable of holding larger values.

It's possible, on the other hand, that if any bits other than the LSB are set in a _Bool, the implementation considers it a trap representation, and it'd be undefined to read from that.

Thanks for all the answers.