为什么我可以在私有类型上使用 auto ?
以下代码编译并运行(vc2012 & gcc4.7.2)让我有些惊讶
I was somehow surprised that the following code compiles and runs (vc2012 & gcc4.7.2)
class Foo {
struct Bar { int i; };
public:
Bar Baz() { return Bar(); }
};
int main() {
Foo f;
// Foo::Bar b = f.Baz(); // error
auto b = f.Baz(); // ok
std::cout << b.i;
}
这段代码编译是否正确?为什么它是正确的?为什么我可以在私有类型上使用 auto，而我不能使用它的名称(如预期的那样)?
Is it correct that this code compiles fine? And why is it correct? Why can I use auto on a private type, while I can't use its name (as expected)?
推荐答案
auto的规则大部分与模板类型推导相同.发布的示例的工作原理与您可以将私有类型的对象传递给模板函数的原因相同:
The rules for auto are, for the most part, the same as for template type deduction. The example posted works for the same reason you can pass objects of private types to template functions:
template <typename T>
void fun(T t) {}
int main() {
Foo f;
fun(f.Baz()); // ok
}
您问，为什么我们可以将私有类型的对象传递给模板函数?因为只有类型的名称是不可访问的.该类型本身仍然可用，这就是为什么您可以将其返回给客户端代码.
And why can we pass objects of private types to template functions, you ask? Because only the name of the type is inaccessible. The type itself is still usable, which is why you can return it to client code at all.
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