c++ - 需要帮助理解返回引用的函数

Question

有什么区别

课本上写的是什么:

MyClass& Myclass::operator++() {
  do something
  return *this;
}

和

MyClass Myclass::operator++() {
  do something
  return *this;
}

* 表示“指向的值”....

是否第二个示例将返回 this 指向的对象的拷贝(*this 的拷贝)，而第一个示例将返回 *this 本身？

如果是这样的话，那有什么不同呢？改善执行时间？

Answer 1

两者的区别在于

情况1

you are returning a reference ie you are returning a constant pointer to the object.

情况2

you are creating a new object and returning the object

让我们深入挖掘以更好地理解事物

情况1

since you are returning the pointer to the existing object any changes you make with affect the original object

情况2

since its a copy there is no impact on the original object

在速度和内存方面

如果是 1

since you returning the address of an existing object there is no or little overhead

情况2

you are creating a new copy ...so constructor of the object would be invoked and hence there is overhead in terms of memory and time taken