了解 C++ 中按值传递对象时的复制构造函数和临时对象创建

问：

给定下一个代码：

#include <iostream>
#include <vector>
#include <cstring>

class Person
{
public:
   Person(const char *i_Name = "Unknown", int i_Age = 0)
       : m_Name(copyFromTo(i_Name, m_Name)), m_Age(i_Age)
   {
      std::cout << "creating obj " << this << ".\n";
   }
   Person(const Person &other)
   {
      std::cout << "Copying from " << &other << " to " << this << ".\n";
      this->m_Name = copyFromTo(other.m_Name, this->m_Name);
      this->setAge(other.m_Age);
   }
   ~Person()
   {
      std::cout << "destroying obj " << this << ".\n";
      if (m_Name)
      {
         delete[] m_Name;
      }
   }
   void printDetails() { std::cout << "Hello, my name is " << m_Name << " and I am " << m_Age << " years old.\n"; }
   void setAge(int i_Age) { m_Age = i_Age; }
   int getAge() const { return m_Age; }

private:
   char *copyFromTo(const char *i_Source, char *i_Destination)
   {
      int sourceSize = strlen(i_Source);
      i_Destination = new char[sourceSize + 1];
      memcpy(i_Destination, i_Source, sourceSize);
      i_Destination[sourceSize] = '\0';
      return i_Destination;
   }
   char *m_Name;
   int m_Age;
};

Person birthday(Person person)
{
   std::cout << "person's address inside the function " << &person << ".\n";
   person.setAge(person.getAge() + 1);
   return person;
}

int main()
{
   Person person = {"New Person", 24};
   birthday(person);
   std::cout << "END OF PROGRAM.\n";
}

下一个输出：

输出：

creating obj 0x61fea8.
Copying from 0x61fea8 to 0x61feb8.
person's address inside the function 0x61feb8.
Copying from 0x61feb8 to 0x61feb0. // [temporary object creation]
destroying obj 0x61feb0. // [temporary object destruction]
destroying obj 0x61feb8.
END OF PROGRAM.
destroying obj 0x61fea8.

请注意，程序会在我们输入函数 'birthday' 后立即创建一个对象，因为参数 'person' 是按值传递的。但是，在函数的末尾，还有对复制构造函数的另一次调用，我们正在创建一个额外的临时对象，然后立即销毁该对象。如果我们在销毁本地“person”对象之前销毁它，为什么还需要这个临时对象，而不是 “birthday” 函数中的本地 'person' 对象。

谢谢。

我尝试了谷歌，看到在函数结束之前创建了一个临时对象，但我不明白如果我甚至在销毁函数的第一个本地对象之前就销毁了它，为什么还需要它。它实际上是用来做什么的？