Question

我最近阅读了简单的STL源代码，以下是代码：

#include <iostream>
#include <cstddef>

using namespace std;

class alloc
{

};

template <typename T, typename Alloc = alloc, size_t Bufsiz = 0>
class deque
{
public:
    deque() { cout << "deque()" << endl; }
};

template <typename T, typename Sequence = deque<T> >
class stack
{
public:
    stack() { cout << "stack()" << endl; }
private:
    Sequence c;
};


int main()
{
    stack<int> x;

    return 0;
}

输出：

deque()
stack()

当我创建一个堆栈对象时，它应该首先调用堆栈构造函数。但事实并非如此。

为什么编译器首先调用deque构造函数？

Answer 1

在输入构造函数的主体之前，会调用基类构造函数，然后所有非静态成员变量都按声明的顺序进行默认初始化，除非它们出现在成员初始化列表中。在您的代码Sequence c首先进行初始化，然后执行Stack::Stack()的正文。

这个程序说明了施工的顺序 - 破坏。

#include <iostream>
#include <string>

using namespace std;

struct Base_1
{
    Base_1()
    {
        cout << "Base_1\n";
    }
    ~Base_1()
    {
        cout << "~Base_1\n";
    }
};

struct Base_2
{
    Base_2()
    {
        cout << "Base_2\n";
    }
    ~Base_2()
    {
        cout << "~Base_2\n";
    }
};

struct Member_1
{
    Member_1()
    {
        cout << "Member_1\n";
    }
    ~Member_1()
    {
        cout << "~Member_1\n";
    }
};

struct Member_2
{
    Member_2()
    {
        cout << "Member_2\n";
    }
    ~Member_2()
    {
        cout << "~Member_2\n";
    }
};

struct Member_non_default
{
    Member_non_default( string s )
    {
        cout << "Member non default\n";
    }
    ~Member_non_default()
    {
        cout << "~Member non default\n";
    }
};

struct Static_member
{
    Static_member()
    {
        cout << "Static member\n";
    }
    ~Static_member()
    {
        cout << "~Static member\n";
    }
};

struct Derived: Base_1, Base_2
{
    Member_1 m1;
    Member_non_default m;
    Member_2 m2;
    static Static_member sm;

    Derived():
        m { "Member non default\n" }
    {
        cout << "Derived\n";
    }
    ~Derived()
    {
        cout << "~Derived\n";
    }
};

Static_member Derived::sm;

int main()
{
    Derived d;
}

输出

静态成员
  Base_1
  Base_2
  MEMBER_1
  会员非默认
  MEMBER_2
  派生
  〜衍生
  〜MEMBER_2
  〜会员非默认
  〜MEMBER_1
  〜Base_2
  〜Base_1
  〜静态成员

STL中构造函数调用的顺序

1 个答案: