可变参数模板的统一初始化 [英] uniform initialization with variadic templates

问题描述

我有一个POD ChParam ，它是可变参数模板函数 set 中的一个参数.我想在花括号 p.set({Param :: D，1000.f}，{Param :: p，2000.f})中传递给函数形参(构造函数参数).并且认为构造函数将被隐式调用，并且将创建 ChParam 对象.但这是不可能的，我应该显式创建一个对象 a.set(ChParam {Param :: D，1000.f}，ChParam {Param :: p，2000.f});

I have a POD ChParam and it's a parameter in the variadic template function set. I'd like to pass to function arguments(constructor parameters) in curly braces p.set({ Param::D, 1000.f }, { Param::p, 2000.f }). And thought that the constructor will be called implicitly and the ChParam objects will be created. But it's impossible, I should explicitly create an object a.set(ChParam{ Param::D, 1000.f }, ChParam{ Param::p, 2000.f });

是否有可能使用变体 p.set({Param :: D，1000.f}，{Param :: p，2000.f})?

#include <iostream>
using namespace std;

using Float = float;

enum class Param : size_t
{
    D = 0,
    p
};
struct ChParam
{
    Param tag_;
    Float value_;
};
class PipeCalcParams
{
private:
    Float D_, p_;
public:
    PipeCalcParams() : D_(0), p_(0) {}
    PipeCalcParams& set_D(Float D) { D_ = D; return *this; }
    PipeCalcParams& set_p(Float p) { p_ = p; return *this; }


    template <typename... Args>
    PipeCalcParams& set(const ChParam& p, Args&&... args) {
        set(p);
        return set(args...);
    }

    PipeCalcParams& set(const ChParam& p)
    {
        switch (p.tag_)
        {
        case Param::D:
            set_D(p.value_);
            break;
        case Param::p:
            set_p(p.value_);
            break;
        }

        return *this;
    }

};

int main() {
    PipeCalcParams a;
    a.set(ChParam{ Param::D, 1000.f }, ChParam{ Param::p, 2000.f });//OK

    PipeCalcParams p;
    p.set({ Param::D, 1000.f }, { Param::p, 2000.f });//error: no matching function for call to 'PipeCalcParams::set(<brace-enclosed initializer list>, <brace-enclosed initializer list>)' p.set({ Param::D, 1000.f }, { Param::p, 2000.f });
    return 0;
}

推荐答案

不能直接使用

{ Param::D, 1000.f }

作为需要推论的函数参数.这样做的原因是初始化列表列表没有类型.由于没有类型，因此编译器无法推断类型.您必须一直提供帮助.您可以执行您的操作并指定类似的类型

as a function parameter when it needs to be deduced. The reason for this is a braced initializer list has no type. Since it does not have a type, a type cannot be deduced by the compiler. You have to help it along. You can do what you did and specify the type like

ChParam{ Param:D, 1000.f }

或者您可以指定所需的对象类型.如果要使用相同类型的变量号，请使用 std :: intializer_list 会起作用.它允许编译器从单独的支撑式初始化列表中构造元素.使用该代码看起来像

Or you can specify the type of object you are expecting. If you want a variable numbers of the same types then a std::intializer_list will work. It allows the compiler to construct the elements from the individual braced initializer lists. Using that your code would look like

PipeCalcParams& set(std::initializer_list<ChParam> args)

当您调用它时，您将使用

And when you call it you would use

p.set({{ Param::D, 1000.f }, { Param::p, 2000.f }})

请注意使用的花括号的额外集合.最外面的集合声明 std :: intializer_list ，每个内部集合声明列表中的每个 ChParam .

Do note the extra set of curly braces used. The outermost set declares the std::intializer_list and each inner set declares each ChParam in the list.

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