C++ 中的用户定义转换

发布于 2024-09-05 04:17:24 字数 750 浏览 4 评论 0原文

最近，我正在浏览 O'Reilly Media 的《C++ Pocket Reference》副本，当我看到有关用户定义类型的用户定义转换的简短部分和示例时，我感到很惊讶：

#include <iostream>

class account {

    private:
        double balance;

    public:
        account (double b) { balance = b; }

        operator double (void) { return balance; }
};

int main (void) {

    account acc(100.0);
    double balance = acc;

    std::cout << balance << std::endl;

    return 0;
}

我已经用 C++ 编程有一段时间了，这是我第一次看到这种运算符重载。这本书对这个主题的描述有些简短，给我留下了一些关于这个功能未解答的问题：

这是一个特别晦涩的功能吗？正如我所说，我用 C++ 编程已经有一段时间了，这是我第一次遇到这种情况。我没有太多运气找到与此相关的更深入的材料。
这个相对便携吗？（我正在 GCC 4.1 上进行编译）
可以完成用户定义类型到用户定义类型的转换吗？例如
operator std::string () { /* code */ }

原文

Recently, I was browsing through my copy of the C++ Pocket Reference from O'Reilly Media, and I was surprised when I came across a brief section and example regarding user-defined conversion for user-defined types:

#include <iostream>

class account {

    private:
        double balance;

    public:
        account (double b) { balance = b; }

        operator double (void) { return balance; }
};

int main (void) {

    account acc(100.0);
    double balance = acc;

    std::cout << balance << std::endl;

    return 0;
}

I've been programming in C++ for awhile, and this is the first time I've ever seen this sort of operator overloading. The book's description of this subject is somewhat brief, leaving me with a few unanswered questions about this feature:

Is this a particularly obscure feature? As I said, I've been programming in C++ for awhile and this is the first time I've ever come across this. I haven't had much luck finding more in-depth material regarding this.
Is this relatively portable? (I'm compiling on GCC 4.1)
Can user-defined conversions to user defined types be done? e.g.
operator std::string () { /* code */ }

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梦忆晨望 2024-09-12 04:17:24

这是一个特别晦涩的功能吗？

是的，转换运算符并不经常使用。我见过它们的地方是用户定义的类型，可以降级为内置类型。像支持与原子数类型之间转换的固定精度数字类之类的东西。

这相对便携吗？

据我所知，确实如此。它们永远处于标准之中。

可以进行用户定义类型到用户定义类型的转换吗？

是的，这是构造函数的功能之一。采用单个参数的构造函数有效地创建从参数类型到类类型的转换运算符。例如，像这样的类：

class Foo {
public:
    Foo(int n) {
        // do stuff...
    }
}

让你这样做：

Foo f = 123;

如果您以前使用过 std::string，那么您很可能在没有意识到的情况下使用过此功能。（顺便说一句，如果您想防止这种行为，请使用显式声明任何单参数构造函数。）

Is this a particularly obscure feature?

Yes, conversion operators aren't used very often. The places I've seen them are for user-defined types that can degrade to built-in ones. Things like a fixed-precision number class that supports converting to/from atomic number types.

Is this relatively portable?

As far as I know, it is. They've been in the standard forever.

Can user-defined conversions to user defined types be done?

Yes, that's one of the features of constructors. A constructor that takes a single argument effectively creates a conversion operator from the argument type to your class's type. For example, a class like this:

class Foo {
public:
    Foo(int n) {
        // do stuff...
    }
}

Let's you do:

Foo f = 123;

If you've used std::string before, odds are you've used this feature without realizing it. (As an aside, if you want to prevent this behavior, declare any single-argument constructors using explicit.)

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