赋值运算符重载：返回 void 与返回引用参数 [duplicate]

问：

我在网上看到的一些赋值重载运算符示例如下所示：

#include <iostream>
using namespace std;

class Distance {
   private:
      int feet;             // 0 to infinite
      int inches;           // 0 to 12
   public:
      // required constructors
      Distance(){
         feet = 0;
         inches = 0;
      }

      Distance(int f, int i){
         feet = f;
         inches = i;
      }

      void operator = (const Distance &D ) { 
         cout << "assigning..." << endl;
         feet = D.feet;
         inches = D.inches;
      }

      // method to display distance
      void displayDistance() {
         cout << "F: " << feet <<  " I:" <<  inches << endl;
      }

};

int main() {
   Distance D1(11, 10), D2(5, 11);

   cout << "First Distance : "; 
   D1.displayDistance();
   cout << "Second Distance :"; 
   D2.displayDistance();

   // use assignment operator
   D1 = D2;
   cout << "First Distance :"; 
   D1.displayDistance();

   return 0;
}

它们从重载函数返回 void。如果 D1 是被调用的对象，这对我来说是有意义的。

其他示例返回对类对象的引用。

#include <iostream>
using namespace std;

class Distance {
   private:
      int feet;             // 0 to infinite
      int inches;           // 0 to 12
   public:
      // required constructors
      Distance(){
         feet = 0;
         inches = 0;
      }

      Distance(int f, int i){
         feet = f;
         inches = i;
      }

      Distance& operator = (const Distance &D ) { 
         cout << "assigning..." << endl;
         feet = D.feet;
         inches = D.inches;
         return *this;
      }

      // method to display distance
      void displayDistance() {
         cout << "F: " << feet <<  " I:" <<  inches << endl;
      }

};

int main() {
   Distance D1(11, 10), D2(5, 11);

   cout << "First Distance : "; 
   D1.displayDistance();
   cout << "Second Distance :"; 
   D2.displayDistance();

   // use assignment operator
   D1 = D2;
   cout << "First Distance :"; 
   D1.displayDistance();

   return 0;
}

这对我来说没有意义（考虑到第一个例子）。如果在第一个示例中调用类似 的东西，为什么第二个示例在这种情况下会起作用？是不是有点像？归根结底，这有什么不同吗？D1 = D2;D1.=(D2);D1 = D1.=(D2);