从const成员函数返回非const引用 [英] Returning non-const reference from a const member function

问题描述

为什么返回对指向的成员变量的引用工作，而不是其他变量工作？我知道一个 const 成员函数应该只返回 const 引用，但为什么这不似乎真的指针？ / p>

Why does returning the reference to a pointed-to member variable work, but not the other? I know that a const member function should only return const references, but why does that not seem true for pointers?

class MyClass
{
  private:
    int * a;
    int b;
  public:
    MyClass() { a = new int; }
    ~MyClass() { delete a; }

    int & geta(void) const { return *a; } // good?
    int & getb(void) const { return b; }  // obviously bad
};

int main(void)
{
  MyClass m;

  m.geta() = 5;  //works????
  m.getb() = 7;  //doesn't compile

  return 0;
}

推荐答案

int & geta(void) const { return *a; } // good?
int & getb(void) const { return b; }  // obviously bad

在const函数中，每个数据成员变为const 无法修改的方式。 int 变为 const int ， int * $ c> int * const ，等等。

In a const-function, every data member becomes const in such way that it cannot be modified. int becomes const int, int * becomes int * const, and so on.

由于您的第一个函数中的 a 的类型变为 int * const ，而不是 const int * ，因此您可以更改数据（可修改）：

Since the type of a in your first function becomes int * const, as opposed to const int *, so you can change the data (which is modifiable):

  m.geta() = 5;  //works, as the data is modifiable

区别： const int * 和 int * const 。

• const int * 表示指针为非常量，但指针指向的数据为 const 。


• int * const 表示指针 const ，但指针指向的数据是 / strong>。

• const int* means the pointer is non-const, but the data the pointer points to is const.
• int * const means the pointer is const, but the data the pointer points to is non-const.

您的第二个函数尝试返回 const int& ，因为 b 的类型变为 const int 。但是你在代码中提到了 int& 的实际返回类型，所以这个函数不甚至编译（参见这个），不管你在 main（）中做什么，因为返回类型不匹配。这里是修复：

Your second function tries to return const int &, since the type of b become const int. But you've mentioned the actual return type in your code as int &, so this function would not even compile (see this), irrespective of what you do in main(), because the return type doesn't match. Here is the fix:

 const int & getb(void) const { return b; }  

现在 compiles fine！。

这篇关于从const成员函数返回非const引用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！