In the below code, I have created a very basic template class Pair and its 2 objects - p1 & p2.

I can get the value of data member x of object p2 using cout << p2.getx().getx() << p2.getx().gety();.

在下面的代码中，我创建了一个非常基本的模板类对及其2个对象-p1和p2。我可以使用cout<

Main Question:

If I can GET the value in this way, then why cannot I SET the value of x using
(p2.getx()).setx('c');
and
(p2.getx()).sety(8.12); ?

如果我可以用这种方式获得值，那么为什么不能使用(p2.getx()).setx(‘c’)；和(p2.getx()).sty(8.12)；来设置x的值呢？

Code:

#include<iostream>
using namespace std;

template <typename T, typename V> class Pair {
  T x;
  V y;

  public:
  void setx(T x) {
    this->x = x;
  }
  T getx() {
    return x;
  }
  void sety(V y) {
    this->y = y;
  }
  V gety() {
    return y;
  }
};

int main()
{
  Pair <char,int> p1;  //Declared to put in p2
  p1.setx('b');
  p1.sety(12.23);
  
  Pair <Pair<char,int>, double> p2;
  p2.setx(p1);
  p2.sety(24.36);

  cout<<p2.getx().getx()<<" "<<p2.getx().gety()<<" "<<p2.gety()<<endl;
  (p2.getx()).setx('c');  //Here, p2.getx()!=p1, even though we set p2.x=p1. This is because p2.getx() returns a value/copy of p1 (an instance), not an object (or var).
  (p2.getx()).sety(8.12);
  p2.setx(p1);
  cout<<p2.getx().getx()<<" "<<p2.getx().gety()<<" "<<p2.gety()<<endl;
  return 0;
}

What I think I've understood so far is that p2.getx() returns a copy of p1, and to make p2.getx().setx(value) work, I would have to make it return a reference to p1.

But still what I would like to understand is that if p2.getx().𝒔𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆) doesn't work, how does p2.getx().𝒈𝒆𝒕𝒙() still work?

我认为到目前为止我已经理解的是，p2.getx()返回p1的副本，要使p2.getx().setx(值)工作，我必须让它返回对p1的引用。但我仍然想了解的是，如果p2.getx().𝒔𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆)不工作，p2.getx().𝒈𝒆𝒕𝒙()如何工作呢？

The p2.getx().𝒈𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆) works because you're returning a copy.
Let's say that I have a function that accept an integer value by copy int functionX(int x) and increments the value that it receives with ++x and I'm passing an int value_int = 50 to the function.
The body of functionX will be able to read x as 50 because it's a copy of value_int. If you runstd::cout << functionX and thenstd::cout << value_int you'll find out that the first line will print 51 and the second one 50, that basically means that passing by copy creats problems only with the writing of an object not the reading. If you want to change values of an object inside a function you should pass it by reference and in your example the first function should return a reference because your're chaining two calls

P2.getx().𝒈𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆)可以工作，因为您要返回一个副本。假设我有一个函数，它通过复制int函数X(Intx)接受一个整数值，并用++x递增它接收到的值，我将一个int value_int=50传递给该函数。函数X的主体将能够将x读取为50，因为它是Value_int的副本。如果运行std：：cout<

For your main question, your getx() method returns a copy of x inside p2, if you set value to this copy, will not affect the x inside p2. If you want to change the x inside p2, your getx() method should return a reference like this:

对于您的主要问题，您的getx()方法返回p2内的x的副本，如果您为该副本设置值，则不会影响p2内的x。如果要更改p2中的x，则getx()方法应返回如下引用：

T& getx() { return x; }

For this question: "But still what I would like to understand is that if p2.getx().𝒔𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆) doesn't work, how does p2.getx().𝒈𝒆𝒕𝒙() still work?"

对于这个问题：“但我仍然想了解的是，如果p2.getx().𝒔𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆)不工作，p2.getx().𝒈𝒆𝒕𝒙()怎么还能工作呢？”

p2.getx().𝒔𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆) doesn't work because you didn't set the x inside p2 directly as I explain before. p2.getx().𝒈𝒆𝒕𝒙() works because the first getx() returns a copy of x inside p2, and the second getx() will call the getx() of the copy, so it works.

P2.getx().𝒔𝒆𝒕𝒙(𝒗𝒂𝒍𝒖𝒆)不起作用，因为您没有像我前面解释的那样直接在p2中设置x。P2.getx().𝒈𝒆𝒕𝒙()之所以起作用，是因为第一个getx()返回了p2内的x的一个副本，而第二个getx()将调用该副本的getx()，所以它起作用了。

Hope I clarify clearly.

希望我能说清楚。