What is the point in expression: !!var ?

[Cedric Blancher]

The point is to transform the value of a (any kind, like pointer) to a
Boolean value.

Ancient C trick, goes back to dmr

Ced

On 13 September 2016 at 21:54, twilight <pipfstarrd at openmailbox.org> wrote:
> Hello,
>
> I'm digging through FreeBSD cppcheck scan report with hope to catch some
> bugs.
>
> I've came along a line:
>         c = cond(token, val, yylex(), noeval | !!a);
>
> What's the point in using !! twice? Can it be deleted?
> Here is the context:
>
> static arith_t cond(int token, union yystype *val, int op, int noeval)
> {
>         arith_t a = or(token, val, op, noeval);
>         arith_t b;
>         arith_t c;
>
>         if (last_token != ARITH_QMARK)
>                 return a;
>
>         b = assignment(yylex(), noeval | !a);
>
>         if (last_token != ARITH_COLON)
>                 yyerror("expecting ':'");
>
>         token = yylex();
>         *val = yylval;
>
>         c = cond(token, val, yylex(), noeval | !!a);
>
>         return a ? b : c;
> }
>
>
> --
> Cheers~
>
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-- 
Cedric Blancher <cedric.blancher at gmail.com>
[https://plus.google.com/u/0/+CedricBlancher/]
Institute Pasteur