auto

[Ary Borenszweig]

Ary Borenszweig wrote:
> bearophile wrote:
>> Ary Borenszweig:
>>> Since when a non-templated function can have its return type deduced? 
>>> :)<
>>
>> More than two months ago I have asked for such feature. Then I have 
>> tested it, and it was already present :-) So I think it's OK, and 
>> expected, it helps when you return very complex types (for example 
>> lazy templated generators).
> 
> Ah... but it wasn't mentioned anywhere.
> 
> I was asking because I get parser errors with Descent if I write that. 
> I'm trying to see where I made a mistake in porting the parser, but the 
> code for parsing that is:
> 
> ---
> /* Look for return type inference for template functions.
>      */
> {
>     Token *tk;
>     if (storage_class &&
>     token.value == TOKidentifier &&
>     (tk = peek(&token))->value == TOKlparen &&
>     skipParens(tk, &tk) &&
>     peek(tk)->value == TOKlparen)
>     {
>         ts = NULL;
>     }
>     else
>     {
>         ts = parseBasicType();
>         ts = parseBasicType2(ts);
>     }
> }
> ---
> 
> and there's also a similar code somewhere else. The parser is there 
> after consuming the "auto" token, so token's value is TOKidentifier. As 
> you can see, it checks that the next token is "(", then it peeks past 
> the closing parenthesis and checks that that token is also "(" (for a 
> templated function). So it won't work with non-templated functions. And 
> parseBasicType and parseBasicType2 fail.
> 
> I'll probably need to debug the C++ code... unless someone can debug it 
> for me and tell me what's going on. :)

Heh, debugging the code revealed that also this condition is present:

|| peek(tk)->value == TOKlcurly

so that's why it parses it ok. Ah... mistery solved.