有这样的代码:对象从功能和拷贝构造函数返回
#include <iostream>
class A {
public:
int a;
A() : a(0) {
std::cout << "Default constructor" << " " << this << std::endl;
}
A(int a_) : a(a_) {
std::cout << "Constructor with param " << a_ << " " << this << std::endl;
}
A(const A& b) {
a = b.a;
std::cout << "Copy constructor " << b.a << " to " << a << " " << &b << " -> " << this << std::endl;
}
A& operator=(const A& b) {
a=b.a;
std::cout << "Assignment operator " << b.a << " to " << a << " " << &b << " -> " << this << std::endl;
}
~A() {
std::cout << "Destructor for " << a << " " << this << std::endl;
}
void show(){
std::cout << "This is: " << this << std::endl;
}
};
A fun(){
A temp(3);
temp.show();
return temp;
}
int main() {
{
A ob = fun();
ob.show();
}
return 0;
}
结果:
Constructor with param 3 0xbfee79dc
This is: 0xbfee79dc
This is: 0xbfee79dc
Destructor for 3 0xbfee79dc
目标对象OB是由函数fun()初始化。为什么复制构造函数没有在那里调用?我认为,当函数按值返回时,将调用复制构造函数或赋值运算符。看起来函数fun()中构造的对象在执行函数后不会被销毁。在这种情况下,拷贝构造函数如何被强制调用?
这是由g ++编译的。