C++ Overloading takes precedence over Specialization?
c++, overload-resolution, overloading, template-specialization, templates
Solution
This is because non-template functions are first-class citizens. See this article by Herb Sutter or this SO post for details.
From Herb Sutter's article:
Nontemplate functions are first-class citizens. A plain old nontemplate function that matches the parameter types as well as any function template will be selected over an otherwise-just-as-good function template.
If there are no first-class citizens to choose from that are at least as good, then function base templates as the second-class citizens get consulted next. Which function base template gets selected depends on which matches best and is the "most specialized" (important note: this use of "specialized" oddly enough has nothing to do with template specializations; it's just an unfortunate colloquialism) according to a set of fairly arcane rules:
Problem
Given the following code: ``` #include <iostream> using namespace std; template<typename T> void Print(T t) { cout << t << endl; } template<> void Print<int>(int t) { cout << "int = " << t << endl; } void Print(int i) { cout << "int2 = " << i << endl; } int _tmain(int argc, _TCHAR* argv[]) { Print(1.3); Print("tese"); Print(2); char c; cin >> c; return 0; } ``` Why is the call Print(2) not ambiguous, but instead calling void Print(int i) ? ps: Tested with bcc64.exe and cl.exe.