空结构背后的目的是什么? [英] The purpose behind empty struct?
问题描述
C ++标准库中对auto_ptr的声明
Declarations of auto_ptr from C++ Standard Library
namespace std {
template <class Y> struct auto_ptr_ref {};
template <class X>
class auto_ptr {
public:
typedef X element_type;
// 20.4.5.1 construct/copy/destroy:
explicit auto_ptr(X* p =0) throw();
auto_ptr(auto_ptr&) throw();
template <class Y> auto_ptr(auto_ptr<Y>&) throw();
auto_ptr& operator=(auto_ptr&) throw();
template <class Y> auto_ptr& operator=(auto_ptr<Y>&) throw();
auto_ptr& operator=(auto_ptr_ref<X>) throw();
~auto_ptr() throw();
// 20.4.5.2 members:
X& operator*() const throw();
X* operator->() const throw();
X* get() const throw();
X* release() throw();
void reset(X* p =0) throw();
// 20.4.5.3 conversions:
auto_ptr(auto_ptr_ref<X>) throw();
template <class Y> operator auto_ptr_ref<Y>() throw();
template <class Y> operator auto_ptr<Y>() throw();
};
}
我不明白这部分的目的:
I don't understand the purpose of this part:
template <class Y> struct auto_ptr_ref {};
不声明任何变量,这些变量如何有效:
Without declaring any variable, how can these be valid:
auto_ptr& operator=(auto_ptr_ref<X>) throw();
还有这些:
auto_ptr(auto_ptr_ref<X>) throw();
template <class Y> operator auto_ptr_ref<Y>() throw();
,还有(我只是注意到)我不明白在最后两行中如何使用运算符".语法是否不像"return-type operatoroperand;"那样,返回类型在哪里?操作数?
and also (I just notice) I don't understand how the "operator" are used for the last two lines. Isn't the syntax something like "return-type operatoroperand;", where is the return type? operand?
推荐答案
Google search for "auto_ptr_ref" reveals this detailed explanation.
我还不太明白,但下面的内容看起来像是.没有这种技巧,您可以将auto_ptr
传递到一个函数中,该函数将获得对象的所有权并将变量分配给空指针.通过上面的附加类技巧,在这种情况下,您将得到一个编译错误.
I don't quite get that explanation, but looks like it is for the following. Without that trick you could pass an auto_ptr
into a function that would get ownership of the object and assign your variable to a null pointer. With the extra class trick above you will get a compile error in such case.
这篇关于空结构背后的目的是什么?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!