#include <iostream>
#include <string.h>
using namespace std;
class withCC
{
public:
withCC() {}
withCC(const withCC&) {
cout<<"withCC(withCC&)"<<endl;
}
};
class woCC
{
enum {bsz = 100};
char buf[bsz];
public:
woCC(const char* msg = 0) {
memset(buf, 0, bsz);
if(msg) strncpy(buf, msg, bsz);
}
void print(const char* msg = 0)const {
if(msg) cout<<msg<<":";
cout<<buf<<endl;
}
};
class composite
{
withCC WITHCC;
woCC WOCC;
public:
composite() : WOCC("composite()") {}
void print(const char* msg = 0) {
cout<<"in composite:"<<endl;
WOCC.print(msg);
}
};
int main()
{
composite c;
c.print("contents of c");
cout<<"calling composite copy-constructor"<<endl;
composite c2 = c;
c2.print("contents of c2");
}
运行结果如下:
$ ./a.out
in composite:
contents of c:composite()
calling composite copy-constructor
withCC(withCC&)
in composite:
contents of c2:composite()
而且我不明白为什么
withCC(withCC&)
作为输出的一部分给出。我想composite c2 = c;
导致复制构造函数被执行。但是,如下所示,WITHCC
是class composite
的一部分,为什么要调用它来处理此复制构造函数?谢谢! 最佳答案
之所以调用复制构造函数withCC(withCC&)
是因为composite
的默认复制构造函数将调用其成员数据的所有复制构造函数。并且由于在withCC
类中有一个composite
对象作为成员数据,因此将调用复制构造函数withCC(withCC&)
。