Re: function pointer as template parameter + type deduction

* er:


What's the point of the unnamed template parameter?

One suspects a case Evil Premature Optimization, that the intention is to shave
a (conjectured but quite possibly not even real) nano-second by calling some
routine "directly" instead of having it passed as a routine pointer argument.

EPO is unfortunately a root cause of so much extreme and unnecessary complexity.


You have defined g with two template parameters, one which is anonymous and
therefore cannot ever be deduced, hence must always be explicitly specified.


Don't. :-)

You're into the second root cause of extreme and unnecessary complexity, namely
the Elegant Notation Fetish (ENF), where almost any absurdly huge baggage of
cryptic, complex, counter-intuitive code is deemed acceptable to shave /one/
character, or perhaps two, in a single very unimportant expression somewhere.

But if you strongly feel that providing the type parameter is very un-elegant,
that it simply must be deduced from the specified function, then perhaps like

<code>
struct A{};
void f(const A&a){}

struct F
{
     typedef A ArgType;
     static void effect( ArgType const& a ) { ::f( a ); }
};

template< class Func >
void g( typename Func::ArgType const& t ) {}

int main()
{
     A a;
     g<F>(a);
}
</code>

Cheers & hth.,

- Alf

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