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

Question

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);

};

@MarekR I am porting a library that I didn't write myself from Python to C++. I'd like that the Python and C++ code look, except from low-level classes, such as this one, more or less the same. — Felix
– Felix, Commented Jun 6, 2022 at 14:59

Jarod42 · Accepted Answer · 2022-06-06 14:58:30Z

7

Deduction guide (c++17) might help:

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

Demo.

answered Jun 6, 2022 at 14:58

Jarod42

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Comments

PaulMcKenzie · Accepted Answer · 2022-06-06 15:13:16Z

0

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

answered Jun 6, 2022 at 15:13

PaulMcKenzie

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