Re: Using operator ! twice

["Pedro de Medeiros" <pedrovmm@gmail.com>]

On 6/29/07, Luong Ngo <luong.ngo@gmail.com> wrote:
> Hello,
>
> 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? this seems to be the same
> for me if we do this in math:  -(-( -8)) = -8;


The reason for using double !s is to evaluate something as a
boolean value. I don't know why use it alone in "if ( !! num )",
but consider this code instead:

char *options[] = { "off" , "on" };

printf("Option is %s.\n", options[!!num]);

Since !!num always evaluates to 0 or 1, you don't have to
test it.


Cheers,
Pedro.

-- 
Pedro de Medeiros - Ciência da Computação - Universidade de Brasília
Home Page: http://www.nonseq.net - Linux User No.: 234250
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