Forcing string literal argument to cause std::string class template deduction-CodePudding

I would like to write a class template, which is able to hold different types. However, I want to avoid specializations of type char* or char[]. Character strings should always be std::string. The code below is what I have so far. However, writing Scalar("hello") yields T = char[6] and not T = std::string. I could write Scalar<std::string>("hello") or Scalar(std::string("hello")) to work around the problem. My question is, is there a way to modify the code below such that writing Scalar("hello") works as intended and the class still has only one template parameter?

template <typename T>
class Scalar {
public:

explicit Scalar(const T& val);

};

CodePudding user response：

Deduction guide (c 17) might help:

template <std::size_t N>
Scalar(const char(&)[N]) -> Scalar<std::string>;

Demo.

CodePudding user response：

Another solution is to use std::string_literals:

#include <string>

template <typename T>
class Scalar {
public:
    explicit Scalar(const T& val) {}
};

int main()
{
   using namespace std::string_literals;    
   Scalar test("hello"s);
}

Of course, the use should already be aware to use the literal "s to convert the string-literal to std::string