我可以确定右值引用的类型吗？

Question

I have a serialize function that performs differently based on the type. I'd like to be able to call it with both f(x) and f(5) , but f(5) fails with error No matching function for call to 'f', Candidate function [with T = int] not viable: expects an l-value for 1st argument. If I change f(T& t) to f(T&& t) then f(x) is not arithmetic. How can I recognize both f(x) and f(5) to be arithmetic, and similarly for any type such as the string type below? I don't want to force the input to be const because I want to alter it otherwise.

template<typename T>
void f(T& t)
{
    if constexpr (std::is_arithmetic_v<T>)
    {
        // do stuff
    }
    else if constexpr (std::is_same_v<T, std::string>)
    {
        // do other stuff
    }
    else
    {
        //alter non-const input
    }
}

int main()
{
    int x;
    f(x);
    f(5);
    return 0;
}

Answer 1

You can use a forwarding reference, T&& , to take the argument by either lvalue- or rvalue reference depending on what's passed in.

In case of lvalue, T = int& , so we need to use std::decay_t to remove the reference from the type.

When we pass an rvalue, T = int and decay does nothing.

#include <type_traits>
#include <string>
#include <iostream>

template<typename T>
void f(T&& t)
{
    using T_Type = std::decay_t<T>; // remove const/reference from the type
    if constexpr (std::is_arithmetic_v<T_Type>)
    {
        // do stuff
    }
    else if constexpr (std::is_same_v<T_Type, std::string>)
    {
        // do other stuff
    }
    else
    {
        //alter non-const input
    }
}

int main()
{
    int x;
    f(x);
    f(5);
    return 0;
}

Note that std::decay performs

Applies lvalue-to-rvalue, array-to-pointer, and function-to-pointer implicit conversions to the type T

If any of those cases is not desired, you can use a combination of std::remove_reference and std::remove_const , or in case of c++20 we can use std::remove_cvref .