当一个类中包含其他类的对象时构造函数的调用顺序

1.按在组合类的定义中出现的顺序调用内嵌对象的构造函数（与初始化列表中出现的顺序无关）
2.调用本类构造函数的函数体
组合类构造函数定义的一般形式一般为：
类名::类名（形参表）:内嵌对象1(形参表）,内嵌对象2(形参表）,……
{类的初始化}

#include <iostream>#include <cmath>using namespace std;class point {public:    point(int xx=0,int yy=0)    {        x=xx;        y=yy;        cout<<"create point"<<endl;    }    point(point &p);    int getx(){return x;}    int gety(){return y;}private:    int x,y;};point::point(point &p){    x=p.x;    y=p.y;    cout<<"copy point "<<endl;}class line{public:    line(point xp1,point xp2);    line(line &l);    double getlen(){return len;}private:    point p1,p2;    double len;};line::line(point xp1,point xp2):p1(xp1),p2(xp2){    cout<<"create line"<<endl;    double x=double(p1.getx()-p2.getx());    double y=double(p1.gety()-p2.gety());    len=sqrt(x*x+y*y);}line::line(line &l):p1(l.p1),p2(l.p2){    cout<<"copy line"<<endl;    len=l.len;}int main(int argc, char *argv[]){    point myp1(1,1),myp2(4,5);    line lin(myp1,myp2);    line lin2(lin);    cout<<"length of lin:"<<lin.getlen()<<endl;    cout<<"length of lin2:"<<lin2.getlen()<<endl;    return 0;}

结果如下:
create point
create point
copy point
copy point
copy point
copy point
create line
copy point
copy point
copy line
length of lin:5
length of lin2:5

可以看到在创造line的对象前,先调用了point类的复制构造函数创建了两个point类的对象.之后才开始调用自己的构造函数.