如何使用auto关键字在C ++ 11中返回任意类型？ [英] How to return arbitrary type in C++11 using auto keyword?

问题描述

我有一个类如下所示：

class Container {
    public:
        Container(){
            Doubles["pi"] = 3.1415;
            Doubles["e"] = 2.7182;

            Integers["one"] = 1;
            Integers["two"] = 2;
        }

        // Bracket.cpp:23:9: error: 'auto' return without trailing return type
        // auto& operator[](const std::string&);
        auto& operator[](const std::string& key);

    private:
        std::map<std::string, double> Doubles;
        std::map<std::string, int> Integers;
};

我想重载 operator [] 函数从 Doubles 或整数返回某个值，具体取决于传递的键。但是，如果返回的是 double 或 int $ c>。我想以这种方式实现运算符[] 函数：

I'd like to overload the operator[] function to return something from either Doubles or Integers depending on a key that is passed. However, I don't know a prioi if what will be returned is a double or a int. I'd like to implement the operator[] function in this way:

// Compiler error
// Bracket.cpp:30:1: error: 'auto' return without trailing return type
// auto& Container::operator[](const std::string& key){
auto& Container::operator[](const std::string& key){
    std::cout << "I'm returning the value associated with key: " 
              << key << std::endl;

    auto D_search = Doubles.find(key);
    if (D_search != Doubles.end()){
        std::cout << "I found my key in Doubles with value: " << 
            D_search->second << std::endl;

        return D_search->second;
    }
    else{
        auto I_search = Integers.find(key);
        if (I_search != Integers.end()){
            std::cout << "I found my key in Integers with value: " << 
                I_search->second << std::endl;

            return I_search->second;
        }
        else{
            std::cout << "I didn't find a value for the key." << std::endl;
        }
    }
}

方法来创建单个运算符[] 函数以返回多个类型？

Is there a way to create a single operator[] function to return multiple types?

这个简单的代码：

int main(){

    Container Bucket;

    double pi(Bucket["pi"]);

    std::cout << "The value of pi is: " << pi << std::endl;

    return 0;
 }

推荐答案

auto 作为返回类型只允许推迟返回类型声明，直到声明函数参数之后：

The C++11 version of auto as return type only allows you to postpone the return type declaration until after the function parameters have been declared:

template<typename T1, typename T2>
auto add(T1 x, T2 y) -> decltype(x + y) { return x + y; }

在这里你不能写 decltype（x + y）add ...），因为编译器不知道 x 和 y 是什么。

Here you can't write decltype(x + y) add(...) because the compiler doesn't know what x and y are.

auto 的C ++ 14版本允许编译器返回类型的扣除。它告诉编译器基于body来推导函数的返回类型，但它仍然是一个返回类型，所以它仍然不能做你想要的。

The C++14 version of auto permits return type deduction by the compiler. It tells the compiler to deduce the return type of the function based on the body, but it's still a single return type, so it still does not do what you want.

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