C ++中的常量指针的目的是什么？

问题描述

我是C ++的初学者，对于指针的用途存在以下问题，其中指针的地址或值都无法更改（ constEverything 示例）。

I am a beginner to C++ and have the following question regarding the purpose of pointers where neither the address nor the value that it is pointing to can be changed (constEverything in the example).

示例：

int main()
{
    int i = 1;
    int j = 2;
    int* const constPoint = &i;
    *constPoint = 3; // legal
    constPoint = &j; // illegal
    const int* constVal = &i;
    *constVal = 3; // illegal
    constVal = &j; // legal
    const int* const constEverything = &i;
    *constEverything = 3; // illegal
    constEverything = &j; // illegal
}

在示例中，指针的类型。使用 constVal 您可以更改地址，使用 constPoint 您可以更改基础值。对于 constEverything ，您将无法执行任何操作。

In the example there are different types of pointers. With constVal you can change the address and with constPoint you can change the underlying value. For constEverything you can do neither.

对我来说，这种指针的目的是通过常量引用传递事物。为什么我不应该只使用 const type& val ？对我来说，const引用似乎容易得多，它使 const type * const val 过时。

To me the purpose of such a pointer would be to pass things by constant reference. Why should I not just use const type &val instead? To me a const reference seems a lot easier and it makes const type* const val obsolete.

推荐答案

常量指针很有用，主要有两个原因。

Constant pointers are useful for two principal reasons.

1. this 是 const const 指针/ code>成员函数。

1. this is a const pointer in a const member function.

A const 指针也可以设置为 nullptr 。

A const pointer can also be set to nullptr.

我的第一个观点有点循环，此 可以是引用类型或 const 引用类型，但是引用迟到了C ++标准，并且所有更改都将被取消。

My first point is somewhat circular, this could have been a reference type or a const reference type, but references arrived late into the C++ standard, and any changes would have been breaking.

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