Question

Having the following code:

template<typename T, typename OutStream = std::ostream> struct print {
  OutStream &operator()(T const &toPrint, OutStream &outStream = std::cout) const {
    outStream << toPrint;
    return outStream;
  }
};

This call is erroneous:

print<int>(2);

Error message:

1>main.cpp(38): error C2440: '<function-style-cast>' : cannot convert from 'int' to 'print<T>'
1>          with
1>          [
1>              T=int
1>          ]
1>          No constructor could take the source type, or constructor overload resolution was ambiguous

This call is not erroneous:

print<int> intPrinter;
intPrinter(2);

Can I use a function object somehow without its instantiation?
I cannot use a template function here, because I need partial specialization capabilities.

Answer 1

I think that you want to say

print<int>()(2);

Here, the first parens create a temporary print<int> object by calling the (zero-argument) constructor, then the second parens actually invoke the function call operator on that object. The error you’re getting now is caused by the fact that

print<int>(2);

Is interpreted as a typecast expression to convert 2 into a print<int>, which isn’t what you want (and also isn’t legal).

Hope this helps!