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);
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
(only works if Scalar<T>::Scalar(T)
is not explicit
):
func<int>(a, 2);
1) The default deduction guide is sufficient: demo.