C++ templateの明示的特殊化について

Question

template<int N> void func2(int a = N){
    cout << "Primary:" << a  << endl;
}
template<> void func2<10>(int b){
    cout << "func2<10>:" << b  << endl;
}

int main(void){
    // Your code here!
    func2<100>();
    func2<10>();
}

上記のfunc2<10>()がなぜコンパイルが成功するのか理解できません。

また引数bは10が代入されたことになります。なぜでしょうか？

Answer 1

Hi, @otaky

template<int N> void func2(int a = N) provides default parameter a=N.
template<> void func2<10>(int b) is a specialization of template<int N> void func2(int a = N).

Declaration (and function signature) is determined by the primary template.

Since func2<10>() has no parameter input. It uses default parameter N 10.

Below is a simplified example:

#include <iostream>
using namespace std;

template<int N> void func2(int a = N);

template<> void func2<10>(int b) {
   
    cout << "func2<10>:" << b << endl;
}

int main(void) {
   
    func2<10>();
}

Minxin Yu

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