c and c++ assignment (overloaded = operator)

  • 2020-06-07 05:01:28
  • OfStack

First of all, the various operators in c++ are implemented as functions, such as =, which is the equal sign function.

So when you assign a value to an object with the = sign, you're actually calling the function with the = sign.

Analyze the following code

#include <iostream>
using namespace std;
class Test{
 explicit Test(){
  data = 0;
 explicit Test(int d):data(d){
  cout << "C:" << this << ":"<< this->data << endl;
 // Copy constructor                             
 Test(const Test &t){
  cout << "Copy:" << this << endl;
  data = t.data;
 // overloading = The operator no.                       
 Test& operator= (const Test &t){
  cout << "assign" << this << endl;
  if(this != &t){
   data = t.data;
  return *this;
  cout << "F:" << this << ":" << this->data << endl;
 int data;
int main(){
 Test t1(10);
 Test t2, t3;
 t3 = t2 = t1;
 return 0;

Focus on the following functions

 // overloading = The operator no.                       
 Test& operator = (const Test &t){
  cout << "assign" << this << endl;
  if(this != &t){
   data = t.data;
  return *this;


1. What does operator = mean

2. Why is the parameter a reference type

3. Why is the parameter limited by const

4, Why if(this! = & t) judgment

5. Why is there a return value

6. Why is the return value of type a reference type

Analysis points answer:

Test t2;

t2 = t1; // The actual operation is t2.operator =(t1), so this in the function is t2

1, override the = function of class Test, which will be called when an object of class Test is manipulated with the = sign

2. Avoid calling copy constructors

3. Avoid accidentally modifying the value of the member variable in the parameter t (t.data = 100;)

4. Prevent yourself from assigning values to yourself

5 in order to be able to use t3 = t2 = t1. If there is no return value, then es62EN3.operator =(t2=t1), t2=t1 has no return value, so compile failed.

6, not a reference is ok, using a reference type prevents the old version of the compiler from calling a copy constructor at return, while the new version of the compiler (gcc 4.8.5-20) does not call a copy constructor even without a reference type.


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