Mr_Visionary
Mr_Visionary
2020-04-12 15:07
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为什么会调用拷贝构造函数与析构函数

c++ 课本上的题
代码如下:

include

using namespace std ;
class A
{
int x,y;
public:
A ( ) { cout << "in A's default constructor \n "; }
A ( const A & ) { cout <<"in A's copy constructor \n ";}
~A ( ) {cout <<" in A's destructor \n " ;}
virtual void f() {cout << " in A's f\n " ;}
void g() { cout << " in A's g\n ";}
void h() {f();g();}
};
class B : public A
{ int z;
public:
B(){ cout << " in B's default constructor\n" ; }
B(const B& ) { cout << "in B's copy constructor\n";}
~B ( ){ cout << " in B's destructor\n " ;}
void f ( ) { cout << " in B's f\n " ;}
void g ( ) { cout << " in B's g\n ";}
};

void func1 (A x)
{x.f();
x.g();
x.h();
}
void func2 (A &x)
{x.f();
x.g();
x.h();
}

int main ()
{
cout<<"------Section 1-------\n";
A a;
A *p= new B;
cout<<"------Section 2-------\n";
func1(a);
cout<<"------Section 3-------\n";
func1(*p);
cout<<"------Section 4-------\n";
func2(a);
cout<<"------Section 5-------\n";
func2(*p);
cout<<"------Section 6-------\n";
delete p;
cout<<"------Section 7-------\n";
return 0;
}
为什么在section2 和section3 中调用了A 的拷贝构造函数与析构函数?

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1条回答 默认 最新

  • caozhy
    已采纳

    调用函数传入a的时候,如果不是引用,就需要在堆栈上创建一个a的副本,这个就要拷贝构造函数
    函数退出,就会调用析构函数

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