Question

在C ++中以与对象创建相反的顺序调用析构函数，但我不明白为什么不为对象数组维护它。

#include <iostream>
using namespace std;
class test {
    int nmbr;
    static int c;
public:
    test(int j)
    {
        cout<<"constructor is called for object no :"<<j<<endl;
        nmbr=j;
    };
    ~test()
    {
        c++;
        cout<<"destructor is called for object no :"<<c<<endl;
    };
};

int test::c=0;

int main()
{
  test ob[]={test(1),test(2),test(3)};
  return 0;
}

上述程序输出

constructor is called for object no :1
constructor is called for object no :2
constructor is called for object no :3
destructor is called for object no :1
destructor is called for object no :2
destructor is called for object no :3

但是为什么没有以相反的顺序调用析构函数？

Answer 1

以相反的顺序调用它。你正在打印变量c。看看我在这个节目中的评论 - ＆＃34;我在这里改变了＃34;。您正在打印一个计数变量，您应该打印正在销毁的对象。

您可以在这里阅读更多内容： https://isocpp.org/wiki/faq/dtors#order-dtors-for-arrays

#include <iostream>
using namespace std;
class test {
    int nmbr;
    static int c;
public:
    test(int j)
    {
        cout<<"constructor is called for object no :"<<j<<endl;
        nmbr=j;
    };
    ~test()
    {
        c++;
        // I changed here
        cout<<"destructor is called for object no :"<<nmbr<<endl;
    };
};

int test::c=0;

int main()
{
  test ob[]={test(1),test(2),test(3)};
  return 0;
}

Answer 2

他们是，你的测试错误。在调用析构函数时，您正在访问自己设置的数字。更改析构函数输出以显示类中的实际值，您将自己看到

cout<<"destructor is called for object no :"<<nmbr<<endl;

Answer 3

以构造函数调用的相反顺序调用析构函数。

您的测试是错误地生成和打印值。

尝试使用此代码，您将看到预期的结果。

#include <iostream>

class test {
    int this_instance;
    static int number_of_instances;
public:
    test() : this_instance(number_of_instances++)
    {
        std::cout << "constructor invoked for object :"<< this_instance << '\n';
    };
    ~test()
    {
        std::cout << "destructor invoked for object :" << this_instance << '\n';
    };
};

int test::number_of_instances = 0;

int main()
{
  test first_batch[4];

  return 0;
}

为什么析构函数不会以相反的顺序调用对象数组？

3 个答案: