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- Subject: RES: RES: RES: C++ Typecast - so far newbie question
- From: "SosCpdTerra" <soscpd@...>
- Date: Mon, 2 Jul 2007 17:38:45 -0300
Hell no... that works!!! Thanks a lot Rici. Fixed with figureout(const char
*what, int ever).
Again, thanks.
Rafael
-----Mensagem original-----
De: lua-bounces@bazar2.conectiva.com.br
[mailto:lua-bounces@bazar2.conectiva.com.br] Em nome de Rici Lake
Enviada em: segunda-feira, 2 de julho de 2007 18:24
Para: Lua list
Assunto: Re: RES: RES: C++ Typecast - so far newbie question
On 2-Jul-07, at 3:05 PM, SosCpdTerra wrote:
> What are you talking about? I cannot pass this return value to a
> variable
> and typecast into whatever I want? I guess this is the evil!!
If you want a mutable copy of the string, you need to make a copy of
the string.
>
> Like:
>
> I have this
> void figureout(char *what, int ever)
>
> and I must his 2 args came from lua. Im trying to do this:
>
> int serial(lua_State *L) //my registered function
> {
> int argc = lua_gettop(L);
> const char var1 = lua_tostring(L, 1);
> int var2 = int(lua_tonumber(L, 2));
> figureout(var1, var2);
> return 0;
> }
>
> There is another way? I can't see that way, and that is my problem.
Declare figureout properly:
void figureout(const char *what, int ever)
Functions which take string arguments and do not modify them should not
be declared as taking char *.
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