Re: Using operator ! twice

[Glynn Clements <glynn@gclements.plus.com>]

Luong Ngo wrote:

> I came across several times codes that use ! operator on an integer
> variable. I am curious what is the purpose of using it twice.
> For example:
> 
> int flag;
>                if( !! flag) {
>                    //do something
>                }
> 
> 
> The first ! operator will make !flag to be true if flag is 0, and if
> we not again it become false, which is just exactly the same if we
> just left it as
>    if(flag), since 0 is the same as false. Similar logic for non-0
> value of flag. Then why do we need to use ! operator twice to get back
> the same value as if we don't use at all?

The !! idiom converts anything which can be used as a boolean to an
integer which is either zero or one.

It isn't necessary in a context where any boolean value is acceptable
(e.g. the test of an if, while or do-while statement, or an operand to
the && and || operators), but is useful if you need a value which must
be either zero or one.

> this seems to be the same
> for me if we do this in math:  -(-( -8)) = -8;

However: !(-8) = 0, and !!(-8) = 1.

-- 
Glynn Clements <glynn@gclements.plus.com>