显式模板函数参数规范和C ++ 11中函数参数的隐式转换

Question

Title of the topic is long and cryptic, but question is rather simple.

I am reading 14.8.1 Explicit template argument specification in the latest C++11 Specs draft(N3242=11-0012), page 375

6 Implicit conversions (Clause 4) will be performed on a function argument to convert it to the type of the corresponding function parameter if the parameter type contains no template-parameters that participate in template argument deduction. [ Note: Template parameters do not participate in template argument deduction if they are explicitly specified. For example,

template<class T> void f(T);
class Complex {
  Complex(double);
};
void g() {
  f<Complex>(1); // OK, means f<Complex>(Complex(1))
}

—end note ]

Could someone explain to me, what it trying to say and where is the conversion taking place in the example.
Thanks !

Answer 1

The conversion is taking place right here:

f<Complex>(1);

You are calling a function f that expects a Complex , but you are passing it an int instead. There is a standard conversion from int to double and a user defined conversion from double to Complex .

What the standard is trying to say is that when you explicitly provide template arguments to a template function, those behave as if the function was declared with those types. That is, when you call f<Complex> it behaves as if declared:

void f( Complex );

Otherwise, had the template parameter not being explicitly specified, T would have been deduced to be int and no implicit conversion would have taken place.