使用std :: variant而不是编译时错误导致缺少函数重载时引发异常 [英] Throw exception on missing function overload with std::variant instead of compile time error

问题描述

这是对这个问题

请考虑以下代码

#include <variant>


int add_(int a, int b){
    return a+b;
}

float add_(float a, float b){
    return a+b;
}

float add_(int a, float b){
    return a+b;
}

float add_(float a, int b){
    return a+b;
}

using Number = std::variant<int, float>;

Number add(Number const& lhs, Number const& rhs ){
        return std::visit( []( auto& lhs_, auto& rhs_ )->Number { return {add_( lhs_, rhs_ )}; }, lhs, rhs );
    }

int main(){
    Number a = 1.f;
    Number b = 2.f;

    Number c = add(a, b);
}

通过为数字添加越来越多的类型，并可能具有依赖于更多类型的函数比2个参数要清楚的是，我需要定义很多函数，即在std :: variant中找到的所有可能的参数组合。

By adding more and more types to number and potentially having function that depend on more than 2 arguments, it quickly becomes clear, that I need to define a lot of functions, namely all possible parameter combinations that are found in the std::variant.

但是，并非所有组合都是可能的和/或需要的。如果我尝试调用未针对特定参数组合重载的函数，该如何仅定义所需的函数并引发异常？例如，假设我只想保留add_（int，int）或add_（float，float）函数。

However, not all combinations are possible and/or needed. How can I only define the functions I need and throw an exception if I try to call a function that is not overloaded for the particular parameter combination? For example, let's say I only want to keep the add_(int, int) or add_(float, float) function.

推荐答案

我从此网站

template <class ...Fs>
struct overload : Fs... {
  overload(Fs const&... fs) : Fs{fs}...
  {}

  using Fs::operator()...;
};


Number add(const Number& arg0, const Number& arg1){
    return std::visit(
        overload{
             [](int a, int b)     -> Number{return add_(a, b);}
            ,[](float a, float b) -> Number{return add_(a, b);}
            ,[](auto& a, auto& b) -> Number{throw std::runtime_error("unsupported parameter combination");}
        },
        arg0, arg1
    );
}

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