当参数不同时，重载分辨率不选择模板

Question

Why the third function call of f is not using the function template?

#include <iostream>
using namespace std;

template<class T> void f(T x, T y) { cout << "Template" << endl; }

void f(int w, int z) { cout << "Non-template" << endl; }

int main() {
   f( 1 ,  2 );
   f('a', 'b');
   f( 1 , 'b');
}

Answer 1

Function template type deduction is very stringent. For each pair of function parameter and argument, deduction happens in isolation, and then deduction results are compared:

[temp.deduct.type]

2 In some cases, the deduction is done using a single set of types P and A, in other cases, there will be a set of corresponding types P and A. Type deduction is done independently for each P/A pair, and the deduced template argument values are then combined. If type deduction cannot be done for any P/A pair, or if for any pair the deduction leads to more than one possible set of deduced values, or if different pairs yield different deduced values, or if any template argument remains neither deduced nor explicitly specified, template argument deduction fails.

This means that the type of x is deduced from 1 (and int), and the type of y is deduced from 'b' (a char) as though the other parameter didn't exist. This deduction yields T = int and T = char separately, and since those are different types, deduction must explicitly fail per the above paragraph.

If you want to call an instantiated function, you must specify T explicitly yourself:

f<int>( 1 , 'b');
f<char>( 1 , 'b');

Answer 2

In the third call, the two arguments are of different types, so T cannot be deduced, and the template function is not viable. The non-template one is then the only viable candidate.