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Subject: Re: [boost] [functional] adding overload
From: Nathan Ridge (zeratul976_at_[hidden])
Date: 2012-01-16 17:34:38

> This can be implemented using a suitable function_type that returns
> the function type result_type (arg1_type, ...) from a function type,
> function pointer, function reference, or from a functor which defines
> the types result_type, arg1_type, ... (as for example boost::function
> does).

You can also implement it for any function object with a nontemplated
operator() even if it doesn't provide arg1_type etc. typedefs, by
examining the signature of its operator().


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