如何基于std :: is_same检查返回不同的类型 [英] How to return different types based on std::is_same check

问题描述

考虑以下代码:

template<typename T>
T foo() {
    if (std::is_same<T, int>::value)
        return 5;

    if (std::is_same<T, std::string>::value)
        return std::string("bar");

    throw std::exception();
}

当使用 foo< int>()调用时，会引发错误无法转换'std :: __ cxx11 :: string {aka std :: __ cxx11 :: basic_string< char>}'到'int'作为回报​​.

When called with foo<int>(), it throws an error cannot convert ‘std::__cxx11::string {aka std::__cxx11::basic_string<char>}’ to ‘int’ in return.

我知道解决方案是使用模板专业化，但是我在问是否可以通过 std :: is_same 检查类型来保持当前机制?

I know the solution is to use template specialization, but I'm asking whether it is somehow possible to keep the current mechanism with checking the type via std::is_same?

推荐答案

if 的两个分支在编译时都必须有效，即使其中一个分支从未执行过.

Both branches of an if must be valid at compile-time, even if one of them is never executed.

如果可以访问C ++ 17，请将 if s更改为 if constexpr :

If you have access to C++17, change the ifs to if constexpr:

template<typename T>
T foo() {
    if constexpr (std::is_same<T, int>::value)
        return 5;

    if constexpr (std::is_same<T, std::string>::value)
        return std::string("bar");

    throw std::exception();
}

在C ++ 17之前，您将必须使用模板专门化来对此进行仿真:

Before C++17, you will have to emulate this by using template specialisation:

template<typename T>
T foo()
{
    throw std::exception();
}

template <>
int foo<int>() {
    return 5;
}

template <>
std::string foo<std::string>() {
    return "bar";
}

如果真正的 foo 所做的工作比该示例更多，并且专门化它会导致代码重复，则可以引入一个辅助函数，该函数将仅封装 return /抛出语句，并将其专门化.

If your real foo does more work than this example and specialising it would lead to code duplication, you can introduce a helper function which will encapsulate only the return/throw statements, and specialise that.

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