問題描述

為什么返回對指向的成員變量的引用有效，而另一個無效?我知道 const 成員函數應該只返回 const 引用，但為什么對于指針來說這似乎不是真的?

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;
}

推薦答案

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

在一個 const 函數中，每個數據成員都變成了 const 這樣它就不能被修改.int 變為 const int，int * 變為 int * const，依此類推.

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.

因為在你的第一個函數中 a 的 type 變成了 int * const，而不是 const int *>，因此您可以更改數據(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

區別:const int* 和 int * const.

• const int* 表示指針是非常量，但指針指向的數據是const.
• int * const 表示指針是const，但指針指向的數據是非常量.

• 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.

你的第二個函數試圖返回 const int &，因為 b 的 type 變成了 const int.但是您已經在代碼中將實際返回類型提到為 int &，因此該函數甚至不會 編譯(請參閱 this)，無論您在 main() 中做什么，因為返回的 type 不匹配.這是修復:

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; }  

現在它編譯正常！.

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