C ++隐式转换运算符优先级 | 隐式转换运算符优先级

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问题描述

编辑：按照Mike Seymour的注释，我替换运算符std :: string（）const; code>并相应地更改了实现。这允许隐式转换，但是，由于某种原因，unsigned long int运算符优先于char *运算符，这是不正确的...此外，我不想暴露讨厌的C的东西，如char *外面类，当我有std :: string。我有一个希望，我的CustomizedInt类需要继承一些东西，以支持我想要的功能。任何人请详细说明Mike关于 std :: basic_string 的意见？我不确定我是否理解正确。

Following Mike Seymour's comment, I replaced operator std::string () const; with operator char * () const; and changed the implementation accordingly. This allows implicit casting, but, for some reason, the unsigned long int operator has precedence over the char * operator, which just does not feel right... Also, I don't want to expose nasty C stuff like char * outside the class, when I have std::string. I have a hunch that my CustomizedInt class needs to inherit from some stuff in order to support the feature that I desire. Could anybody please elaborate Mike's comment regarding std::basic_string? I'm not sure I understood it properly.

我有这条代码：

#include <string>
#include <sstream>
#include <iostream>

class CustomizedInt
{
private:
    int data;
public:
    CustomizedInt() : data(123)
    {
    }
    operator unsigned long int () const;
    operator std::string () const;
};

CustomizedInt::operator unsigned long int () const
{
    std::cout << "Called operator unsigned long int; ";
    unsigned long int output;
    output = (unsigned long int)data;
    return output;
}

CustomizedInt::operator std::string () const
{
    std::cout << "Called operator std::string; ";
    std::stringstream ss;
    ss << this->data;
    return ss.str();
}

int main()
{
    CustomizedInt x;
    std::cout << x << std::endl;
    return 0;
}

这里输出Called operator unsigned long int; 123。我的问题是这些：

Which prints "Called operator unsigned long int; 123". My questions are these:

在我删除运算符unsigned long int之后，为什么需要显式地将x转换为std :: string？为什么不直接调用隐式转换运算符（std :: string）？

是否有任何文档解释允许哪些隐式转型，以及它们的优先级顺序？看起来如果我把一个运算符unsigned int和运算符unsigned long int一起添加到这个类中，我会收到一个关于<< operator ...

此外，我知道定义这样的运算符可能是不好的做法，但我不能确定我是否完全理解相关的警告。有人请大概说一下吗？是否更好的做法是只定义公共方法ToUnsignedLongInt和ToString？

推荐答案

; 对于字符串是一个模板，由 std :: basic_string 模板参数（ std :: string 本身是该模板的特殊化）。它只能通过参数相关的查找来选择，并且只有当参数实际上是 std :: basic_string 的专业化，而不是可转换的。

The version of << for strings is a template, parametrised by the parameters of the std::basic_string template (std::string itself being a specialisation of that template). It can only be chosen by argument-dependent lookup, and that only works if the argument is actually a specialisation of std::basic_string, not something convertible to that.

规则相当复杂，你需要阅读完整故事的C ++标准。简单的经验法则是，隐式转换不能包含多个用户定义的转换，并且（如您所知），隐式转换的结果不能用于通过参数相关查找来选择模板专用化。

The rules are quite complex, and you'd need to read the C++ standard for the full story. Simple rules of thumb are that implicit conversions can't contain more than one user-defined conversion and (as you've found out) the result of an implicit conversion can't be used to choose a template specialisation by argument-dependent lookup.

我不完全明白他们吗？隐式转换，名称查找和模板专门化（以及我现在无法想到的其他因素）之间的交互相当复杂，大多数人没有倾向于学习它们。有很多实例，隐式转换不会发生，而其他地方，当你不期望它可能发生;

I don't fully understand them either; the interactions between implicit conversions, name lookup and template specialisation (and probably other factors that I can't think of right now) are rather complex, and most people don't have the inclination to learn them all. There are quite a few instances where implicit conversion won't happen, and others where it might happen when you don't expect it; personally, I find it easier just to avoid implicit conversions most of the time.

这可能是一个好主意，以避免不必要的转换。您可以通过保留它们并在必要时显式使用它们来解决您的问题：

That's probably a good idea, to avoid unwanted conversions. You can fix your problem by leaving them and use them explicitly when necessary:

std::cout << std::string(x) << std::endl;

在C ++ 11中，可以声明 explicit ，因此它们只能以这种方式使用。在我看来，这将是最好的选择，如果你可以;

In C++11, you can declare them explicit, so that they can only be used in this manner. In my opinion, that would be the best option if you can; otherwise, I would use explicit conversion functions as you suggest.

顺便说一下， main（）的返回类型必须 int ，而不是 void 。

By the way, the return type of main() must be int, not void.

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