type defined in free function, accessible through auto outside. Language Bug or Feature?

问题

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;
}

It compiles, run as expected:

0 1

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;
}

What does the standard say about it?

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;
}

回答1:

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.

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.

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

来源：https://stackoverflow.com/questions/33783513/type-defined-in-free-function-accessible-through-auto-outside-language-bug-or