Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
306 views
in Technique[技术] by (71.8m points)

c++ - Why does my destructor appear to be called more often than the constructor?

#include<iostream>
using namespace std;

class A{
public:
    static int cnt;
    A()
    { 
        ++cnt; 
        cout<<"constructor:"<<cnt<<endl;
    }
    ~A()
    {
        --cnt;
        cout<<"destructor:"<<cnt<<endl;
    }
};

int A::cnt = 0;

A f(A x){
    return x;
}
int main(){
    A a0;
    A a1 = f(a0);
    return 0;
}

The program will output:

constructor:1
destructor:0
destructor:-1
destructor:-2

The constructor and destructor don't appear in pairs?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You need to add a copy constructor that increases the counter.

A(const A&)
{ 
    ++cnt; 
    cout<<"copy constructor:"<<cnt<<endl;
}

If you don't add it explicitly, the compiler generates one that does nothing with the counter cnt.

This expression

A a1 = f(a0);

is creating copies of a0, which make use of the copy constructor. The exact number of copies may vary depending on copy elision, but your cnt should be 0 at the end of the program.

Note: In C++11, you should also consider the possibility of a compiler generated move copy constructor, however, once you declare your own copy constructor, the compiler no longer generates the move version.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...