条款19了解临时对象的来源 练习

例1

#include<iostream>using namespace std;class Base{      public:             int a;             int b;             Base(){cout<<"base"<<endl;}             Base(Base &b){cout<<"base(&)"<<endl;}             ~Base(){cout<<"~base"<<endl;}             void mf(){cout<<"B::fun()"<<endl;}};class Derived:public Base{    public:           int c;           Derived(){cout<<"Derived"<<endl;}           Derived(Derived &d){cout<<"Derived(&)"<<endl;}           ~Derived(){cout<<"~Derived"<<endl;}           void mf(){cout<<"D::fun()"<<endl;}};void func(Base& d)//切割，只引用了D类中B的部分。 {      cout<<sizeof(d)<<endl;   d.mf();   cout<<"func"<<endl;}int main(){    Derived d;    func(d);    getchar();    return 0;}

结果：

例2：

#include<iostream>using namespace std;class Base{      public:             int a;             int b;             Base(){cout<<"base"<<endl;}             Base(Base &b){cout<<"base(&)"<<endl;}             ~Base(){cout<<"~base"<<endl;}             void mf(){cout<<"B::fun()"<<endl;}};class Derived:public Base{    public:           int c;           Derived(){cout<<"Derived"<<endl;}           Derived(Derived &d){cout<<"Derived(&)"<<endl;}           ~Derived(){cout<<"~Derived"<<endl;}           void mf(){cout<<"D::fun()"<<endl;}};void func(Derived& d) {      cout<<sizeof(d)<<endl;   d.mf();   cout<<"func"<<endl;}int main(){    Derived d;    func(d);    getchar();    return 0;}

结果：‘

例3：reference-to-const 参数，产生临时变量

#include<iostream>using namespace std;class Base{      public:             int a;             int b;             Base(int x){cout<<"base"<<endl;}             Base(Base &b){cout<<"base(&)"<<endl;}             ~Base(){cout<<"~base"<<endl;}             void mf(){cout<<"B::fun()"<<endl;}};void func(const Base& b) //func(Base& b)则编译不通过 {      cout<<sizeof(b)<<endl;  // b.mf();//编译不通过，b是隐式转换的临时变量时，不能调用Base的成员函数。    cout<<"func"<<endl;}int main(){    int x=9;    func(1);//隐式转换。Int->Base.    getchar();    return 0;}

只有当对象以by value方式传递，或是当对象被传递给一个reference to const参数时，这些转换才会发生。如果对象被传递给一个reference to no-const 参数，并不会发生此类转换。所以上例中func(Base& b)则编译不通过 。

1.任何时候只要看到一个reference-to-const参数，就极可能会产生一个临时对象，绑定至该参数上。

2.任何时候只要你看到一个函数返回一个对象，就会产生临时对象（并于稍后销毁）。

例4：函数返回对象

#include<iostream>using namespace std;class Base{      public:             Base(){cout<<"Base"<<endl;}             Base(Base& b){cout<<"Base(&)"<<endl;}             ~Base(){cout<<"~Base"<<endl;}};Base func(){     return Base();//编译器优化，把对象构造到目标对象上。如同Base tp = Base();}int main(){    func();//相当于const Base temp =  func();temp在函数调用语句结束后即析构。    getchar();    return 0;}

结果：

例5：函数返回对象

#include<iostream>using namespace std;class Base{      public:             Base(){cout<<"Base"<<endl;}             Base(Base& b){cout<<"Base(&)"<<endl;}             ~Base(){cout<<"~Base"<<endl;}};Base func(Base &b){     return b;}int main(){Base ob;    func(ob);//相当于const Base temp =  func(ob);temp在函数调用语句结束后即析构。    getchar();    return 0;}

结果：

例6：函数返回对象

#include<iostream>using namespace std;class Base{      public:             Base(){cout<<"Base"<<endl;}             Base(Base& b){cout<<"Base(&)"<<endl;}             ~Base(){cout<<"~Base"<<endl;}};Base func(Base b){     return b;}int main(){    Base ob;    func(ob);//相当于const Base temp =  func(ob);temp在函数调用语句结束后即析构。    getchar();    return 0;}

结果：

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