在自由功能中定义的类型，可通过外部自动访问。语言错误或功能？ [英] type defined in free function, accessible through auto outside. Language Bug or Feature?

问题描述

让我们在自由函数中定义一个类，然后在外部进行访问：

Let's define a class inside a free function, and access it outside:

#include <iostream>
auto myFunc(){
    class MyType{public: int i = 0; int j = 1;};
    return MyType();
}
int main() {
    auto my_type = myFunc();
    std::cout << my_type.i << " " << my_type.j << "\n";
    return 0;
}

它可以编译，按预期运行：

It compiles, run as expected:

0 1

名称MyType是正确隐藏：
如果我们替换为auto，则不会编译以下内容：

The name MyType is properly hidden: if we replace auto, the following won't compile:

int main() {
    MyType my_type = myFunc();
    std::cout << my_type.i << " " << my_type.j << "\n";
    return 0;
}

标准怎么说？

如何预防？以下代码无济于事：

How to prevent it? The following code did not help:

 namespace{
auto myFunc(){
    class MyType{public: int i = 0; int j = 1;};
    return MyType();
}
}
int main() {
    auto my_type = myFunc();
    std::cout << my_type.i << " " << my_type.j << "\n";
    // your code goes here
    return 0;
}

推荐答案

标准未说具体而言，除了—正如您已经指出的—是具有范围而不是类型的 name 。使用 auto 绕过类型的名称，无论名称范围如何，您都可以使用该类型。

The standard doesn't say anything about this specifically, except that — as you've already pointed out — it's the name that has a scope, not the type. Use of auto bypasses the type's name, giving you a way to get at the type regardless of the name's scope.

有点类似于制作嵌套类 private 并不意味着您不能使用它的实例，只是您不能在封装类的范围之外命名它。

It's kind of similar to how making a nested class private doesn't mean you can't use instances of it, only that you can't name it outside of the encapsulating class's scope.

我不知道您如何d阻止它，也不想这么做。

I don't see how you'd "prevent" it, nor why you'd want to.

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