Topic: decltype(myself)
Author: BetaNona <betanona@gmail.com>
Date: Fri, 7 Aug 2009 12:00:28 CST Raw View
I see. f is not declared at the left of -> .
Thank you.
Thus, is this well-form?
At decltype, f is not declared and so not overloaded.
void *f(int);
auto f() -> decltype(&f) {return 0;}
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Author: Mathias Gaunard <loufoque@gmail.com>
Date: Sat, 8 Aug 2009 09:55:06 CST Raw View
On 7 ao t, 20:00, BetaNona <betan...@gmail.com> wrote:
> Thus, is this well-form?
> At decltype, f is not declared and so not overloaded.
>
> void *f(int);
> auto f() -> decltype(&f) {return 0;}
Yes, this is well formed.
This is equivalent to
void* (*f())(int) { return 0; }
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Author: BetaNona <betanona@gmail.com>
Date: Tue, 4 Aug 2009 13:16:44 CST Raw View
What is the type of f ?
auto f() -> decltype(&f) {return f;}
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Author: Mathias Gaunard <loufoque@gmail.com>
Date: Tue, 4 Aug 2009 19:11:44 CST Raw View
On 4 ao t, 21:16, BetaNona <betan...@gmail.com> wrote:
> What is the type of f ?
>
> auto f() -> decltype(&f) {return f;}
This code is ill-formed.
f is not in scope in decltype(&f).
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Author: Sean Hunt <rideau3@gmail.com>
Date: Wed, 5 Aug 2009 15:19:41 CST Raw View
On Aug 4, 1:16 pm, BetaNona <betan...@gmail.com> wrote:
> What is the type of f ?
>
> auto f() -> decltype(&f) {return f;}
At first glance, that seems ill-formed because f isn't declared until
after the late-specified return type.
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