箭头->运算符重载在C ++内部工作? [英] How arrow-> operator overloading works internally in c++?

问题描述

我了解正常的运算符重载.编译器可以将它们直接转换为方法调用.我对->运算符不太清楚.我正在编写我的第一个自定义迭代器，感觉就像需要->运算符.我看了一下stl源代码，并实现了自己的代码:

I understand the normal operator overloading. Compiler can translate them to method call directly. I am not very clear about the -> operator. I was writing my first custom iterator and I felt like the need of -> operator. I took a look at the stl source code and implemented my own like it:

MyClass* MyClassIterator::operator->() const
{
    //m_iterator is a map<int, MyClass>::iterator in my code.
    return &(m_iterator->second);
}

然后我可以使用MyClassIterator的实例，例如:

Then I can use an instance of MyClassIterator like:

myClassIterator->APublicMethodInMyClass().

看起来编译器在这里执行了两个步骤. 1.调用->()方法以获取一个临时MyClass *变量. 2.使用其->运算符在temp变量上调用APublicMethodInMyClass.

Looks like the compiler does two steps here. 1. Call the ->() method the get a temporary MyClass* variable. 2. Call the APublicMethodInMyClass on the temp variable use its -> operator.

我的理解正确吗?

推荐答案

myClassIterator->APublicMethodInMyClass()

无非是以下内容:

myClassIterator.operator->()->APublicMethodInMyClass()

对重载的operator->的第一次调用会为您提供某种类型的指针，该指针具有一个可访问的(从您的调用站点获取)成员函数，称为APublicMethodInMyClass().当然，根据函数查找规则来解析APublicMethodInMyClass()，具体取决于它是否是虚拟的.

The first call to the overloaded operator-> gets you a pointer of some type which has an accessible (from your call-site) member function called APublicMethodInMyClass(). The usual function look-up rules are followed to resolve APublicMethodInMyClass(), of course, depending on whether it is a virtual or not.

不一定有一个临时变量；编译器可能会也可能不会复制&(m_iterator->second)返回的指针.很有可能，这将被优化掉.不过不会创建类型为MyClass的临时对象.

There is not necessarily a temporary variable; the compiler may or may not copy the pointer returned by &(m_iterator->second). In all probability, this will be optimized away. No temporary objects of type MyClass will be created though.

通常的警告也适用于m_iterator-确保您的调用不会访问无效的迭代器(例如，如果您使用的是vector).

The usual caveats also do apply to m_iterator -- make sure that your calls do not access an invalidated iterator (i.e. if you are using vector for example).

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