了解和使用副本分配构造函数

Question

我试图了解副本分配构造函数在c ++中的工作方式。我只使用过Java，所以我真的不在这里。我已经阅读并看到返回引用是一种很好的做法，但是我不知道该怎么做。我写了这个小程序来测试这个概念：

main.cpp：

#include <iostream>
#include "test.h"

using namespace std;

int main() {
    Test t1,t2;
    t1.setAge(10);
    t1.setId('a');
    t2.setAge(20);
    t2.setId('b');

    cout << "T2 (before) : " << t2.getAge() << t2.getID() << "\n";

    t2 = t1; // calls assignment operator, same as t2.operator=(t1)

    cout << "T2 (assignment operator called) : " << t2.getAge() << t2.getID() << "\n";

    Test t3 = t1; // copy constr, same as Test t3(t1)

    cout << "T3 (copy constructor using T1) : " << t3.getAge() << t3.getID() << "\n";

    return 1;
}

test.h：

class Test {
    int age;
    char id;

    public:
        Test(){};
        Test(const Test& t); // copy
        Test& operator=(const Test& obj); // copy assign
        ~Test();
        void setAge(int a);
        void setId(char i);
        int getAge() const {return age;};
        char getID() const {return id;};
};

test.cpp：

#include "test.h"

void Test::setAge(int a) {
    age = a;
}

void Test::setId(char i) {
    id = i;
}

Test::Test(const Test& t) {
    age = t.getAge();
    id = t.getID();
}

Test& Test::operator=(const Test& t) {

}

Test::~Test() {};

我似乎不明白我应该放什么东西operator=()。我见过有人回来了*this，但从我读到的内容仅是对象本身的引用（在的左侧=），对吗？然后，我想到了返回const Test& t对象的副本，但是使用该构造函数没有意义了吗？我要返回什么，为什么？

Answer 1

我已经阅读并看到返回引用是一种很好的做法，但是我不知道该怎么做。

怎么样

加

return *this;

as the last line in the function.

Test& Test::operator=(const Test& t) {
   ...
   return *this;
}

Why

As to the question of why you should return *this, the answer is that it is idiomatic.

For fundamental types, you can use things like:

int i;
i = 10;
i = someFunction();

You can use them in a chain operation.

int j = i = someFunction();

You can use them in a conditional.

if ( (i = someFunction()) != 0 ) { /* Do something */ }

You can use them in a function call.

foo((i = someFunction());

They work because i = ... evaluates to a reference to i. It's idiomatic to keep that semantic even for user defined types. You should be able to use:

Test a;
Test b;

b = a = someFunctionThatReturnsTest();

if ( (a = omeFunctionThatReturnsTest()).getAge() > 20 ) { /* Do something */ }

But Then

More importantly, you should avoid writing a destructor, a copy constructor, and a copy assignment operator for the posted class. The compiler created implementations will be sufficient for Test.