Explicit template arguments for operator()

[James Daughtry]

It seems impossible to explicitly specify template arguments for an
overloaded operator() without giving the full name. I'd like to know
where in the C++ standard this behavior is defined. I can't seem to
find it. Here's some code the shows what I'm talking about:

#include <iostream>

using namespace std;

class Foo {
public:
template <typename Func>
void operator()() {
cout << "Foo::operator( )\n";
Func()();
}
};

class Bar {
public:
void operator()() {
cout << "Bar::operator( )\n";
}
};

int main() {
Foo f;

// Cannot deduce Func
f();

// Illegal expression
f<Bar>();

// Works fine
f.operator()<Ba r>();
}

[Barry]

Re: Explicit template arguments for operator()

On 9ÔÂ17ÈÕ, ÏÂÎç9ʱ46·Ö, James Daughtry <mordoc...@hotm ail.comwrote:

It seems impossible to explicitly specify template arguments for an
overloaded operator() without giving the full name. I'd like to know
where in the C++ standard this behavior is defined. I can't seem to
find it. Here's some code the shows what I'm talking about:
>

14.8.1/5 give some 'note' closely on this, but not exactly,
I wonder adding something about 'overloaded operators' here is
approriate.

--
Best Regards
Barry