On Wed, Jul 27, 2011 at 7:16 AM, Edward Diener <eldiener@tropicsoft.com>wrote:
On 7/26/2011 5:58 PM, Jeffrey Lee Hellrung, Jr. wrote:
On Tue, Jul 26, 2011 at 2:25 PM, Edward Diener<eldiener@tropicsoft.com**
wrote:
On 7/26/2011 5:03 PM, Jeffrey Lee Hellrung, Jr. wrote:
On Tue, Jul 26, 2011 at 1:19 PM, Noah Roberts<roberts.noah@gmail.com**
**
wrote:
[...]
One thing that I think is missing is the option to check for a named
callable within a type. For example, one might want to use enable_if style concept checking. So long as a type has function 'xxx' that takes arguments of type x,y,and z, and returns something convertible to type A, the concept is obeyed. I don't believe this can be done with C++03, but with C++1X I was able to do something using decltype.
[...]
Actually, you can approximate it pretty well in C++03 (using derivation and using tricks) to the point that I think it will work pretty much as you expect except for nullary member functions, where your queries must be more restrictive. This is a useful query, but I believe Edward has decided it to be outside the scope of TTI.
I am not sure what the OP means by 'named callable". If it can be introspected within the type I can consider whether it can be done or not.
I think you missed the convertibility requirement(s).
TTI needs an exact signature.
...and hence why I had presumed it to be outside of the scope of TTI. It stills falls within the domain of introspection, though.
This is something more along the lines of Frederic Bron's Type Traits
Extension, I think.
That deals with operators for a type in general. I have a feeling that the OP was interested in determining whether a class was a function object for some parameter(s).
Frederic's framework is not really specific to operators, per se. For example, I believe his techniques could be used to detect whether a free function would be found via ADL. The proposed Type Traits Extension is all about "can you call operator X with arguments Y with result convertible to Z?". It's not such a large leap to ask the same question about a member function. - Jeff