1 minute read

整理自:Thinking in C++


有两点意思:

  • T t2 = t1; 这样的 define 语句会调用 copy-constructor(生成一个新对象)
  • t2 = t1; 这样的赋值语句执行的是 operator=
    • Default assigment operator= in c++ is a shallow copy? 的说法是:系统默认的 operator= 执行的是 copy 操作,which copies each member。但个 copy 并不像 copy-constructor 是生成一个新对象,理解为 override 或者 memberwise assignment 似乎更合适
      • t2 = t1; 就是用 t1 的内容 override t2 的内容,或者理解为 t2.mA = t1.mA; t2.mB = t1.mB; ... t2.mZ = t1.mZ;
    • 如果你重载了 operator= 就执行重载的逻辑

我们先来看下 Java 的版本:

public class T {
	private int i;
	
	public int getI() {
		return i;
	}

	public void setI(int i) {
		this.i = i;
	}

	public T(int i) {
		super();
		this.i = i;
	}
	
	public static void main(String[] args) {
		T t1 = new T(47);
		T t2 = t1;
		
		t2.setI(48);
		
		System.out.println(t1.getI());
		System.out.println(t2.getI());
		System.out.println(t1 == t2);
	}
}

// output:
/* 
	48
	48
	true
*/

Java 里都是 reference,所以 t2t1 指向了同一个对象,它们连地址都是相同的。

再来看下 C++ 的版本:

#include <iostream>
using namespace std;

class T {
private:
    int i;
public:
    T(int i);
    T(const T&);
    ~T();
    int getI() const {
        return i; 
    }
    void setI(int i) {
        this->i = i;
    }
};

T::T(int i) {
    this->i = i;
}

T::T(const T& t) : i(t.getI()) {
    cout << "copy-constructor: &t==" << &t << endl;
    cout << "copy-constructor: this==" << this << endl;
} 

T::~T() {
    cout << "destructor: this==" << this << endl;
}

int main() {
    T t1(47);
    T t2 = t1;
    
    t2.setI(48);
    
    cout << "t1.i==" << t1.getI() << endl;
    cout << "t2.i==" << t2.getI() << endl;
    
    T t3(53);
	t2 = t3; 
	
	cout << "t1.i==" << t1.getI() << endl;
	cout << "t2.i==" << t2.getI() << endl;
	cout << "t3.i==" << t3.getI() << endl;
}

// output:
/* 
	copy-constructor: &t==0x22fe30
	copy-constructor: this==0x22fe20
	t1.i==47
	t2.i==48
	t1.i==47
	t2.i==53
	t3.i==53
	destructor: this==0x22fe10
	destructor: this==0x22fe20
	destructor: this==0x22fe30
*/

可见 T t2 = t1; 调用了 copy-constructor,t1t2 是两个完全不同的对象。而 t2 = t3; 是用 t3 的值完全覆盖了 t2t2t3 仍然是两个不同的对象。

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