深奥的C ++运算符 [英] Esoteric C++ operators

问题描述

以下深奥的C ++运算符的作用是什么?

What is the purpose of the following esoteric C++ operators?

指向成员的指针

::*

按指针将指针绑定到成员

Bind pointer to member by pointer

->*

通过引用绑定指向成员的指针

Bind pointer to member by reference

.*

(参考)

推荐答案

指向成员的指针使您可以拥有一个与特定类相关的指针.

A pointer to a member allows you to have a pointer that is relative to a specific class.

所以，假设您有一个包含多个电话号码的联系人课程.

So, let's say you have a contact class with multiple phone numbers.

class contact
{
    phonenumber office;
    phonenumber home;
    phonenumber cell;
};

这个想法是，如果您有一个需要使用电话号码的算法，但决定应该在算法之外使用哪个电话号码，则指向成员的指针可以解决此问题:

The idea is if you have an algorithm that needs to use a phone number but the decision of which phone number should be done outside the algorithm, pointers to member solve the problem:

void robocall(phonenumber contact::*number, ...);

现在，robocall的呼叫者可以确定要使用的电话号码类型:

Now the caller of robocall can decide which type of phonenumber to use:

robocall(&contact::home, ...);    // call home numbers
robocall(&contact::office, ...);  // call office number

一旦有了指针，

.*和->*就会起作用.因此，在robocall内部，您可以这样做:

.* and ->* come into play once you have a pointer. So inside robocall, you would do:

contact c = ...;
c.*number;    // gets the appropriate phone number of the object

或:

contact *pc = ...;
pc->*number;

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