Why does the implicit type conversion not work in template deduction?

Question

In the following code, I want to call a template function by implicitly converting an int to a Scalar<int> object.

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar{
public:
  Scalar(Dtype v) : value_(v){}
private:
  Dtype value_;
};

template<typename Dtype>
void func(int a, Scalar<Dtype> b){ 
  cout << "ok" <<endl;
}

int main(){
  int a = 1;
  func(a, 2); 
  //int b = 2;
  //func(a, b);
  return 0;
}

Why does the template argument deduction/substitution fail? And the commented-codes are also wrong.

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
   func(a, 2);
            ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
      ^
test.cpp:13:6: note:   template argument deduction/substitution failed:
test.cpp:19:12: note:   mismatched types ‘Scalar<Dtype>’ and ‘int’
   func(a, 2);

Answer 1

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>{2}); 

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar{2}); // C++17 only

Alternatively, you can explicitly instantiate f:

func<int>(a, 2); 

---

¹⁾ The default deduction guide is sufficient: demo.

Answer 2

template<typename Dtype>
void func(int a, Scalar<Dtype> b){ 
  cout << "ok" <<endl;
}
template<typename Dtype>
void func(int a, Dtype b){ 
  func(a, Scalar<Dtype>(std::move(b)));
}

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.