C++: STL: vector: remove: 析构函数调用

Question

代码如下:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

struct A {
  A(int i = -1): i_(i) {
    wcout << "Constructor: i = " << i_ << endl;
  }
  A(A const &a) {
    wcout << "Copy constructor: i = " << i_ << " a.i = " << a.i_ << endl;
    *this = a;
  }
  ~A() { wcout << "Destructor: i = " << i_ << endl; }
  A &operator=(A const& a) {
    wcout << "Copy assignment operator: i = " << i_ << " a.i = " << a.i_ << endl;
    i_ = a.i_;
    return *this;
  }
  bool operator==(A const& rhs) { return i_ == rhs.i_; }
  int get() { return i_; }
private:
  int i_;
};

int wmain() {
  A a[] = {1, 2, 3, 2, 4, 5};
  vector<A> v(a, a + sizeof a/sizeof a[0]);
  wcout << "==== Just before remove ====" << endl;
 remove(v.begin(), v.end(), 2);
  wcout << "==== Just after remove ====" << endl;

  return 0;
}

输出:

==== Just before remove ====
Constructor: i = 2
Destructor: i = 2
Constructor: i = 2
Destructor: i = 2
Constructor: i = 2
Destructor: i = 2
Copy assignment operator: i = 2 a.i = 3
Constructor: i = 2
Destructor: i = 2
Constructor: i = 2
Destructor: i = 2
Copy assignment operator: i = 3 a.i = 4
Constructor: i = 2
Destructor: i = 2
Copy assignment operator: i = 2 a.i = 5
==== Just after remove ====

问题是:为什么在运行 remove() 时调用了 6 次析构函数？我需要澄清这个困惑。

注意:请在您的系统上执行此代码，然后再回答。注意:MSVCPP 11

Answer 1

The question is: why destructor is called 6 times while remove() was running?

总而言之，析构函数调用与 2 通过 remove() 隐式转换为 A 有关。每当这种隐式转换的结果超出范围时，A 的析构函数就会被调用。

这些隐式转换的原因是 remove() 需要将 a 的每个元素与 2 进行比较。唯一的方法是调用 A::operator==(const A&):

  bool operator==(A const& rhs) { ... }

由于 rhs 是 const A& 类型，编译器:

1. 调用A(int)将2转换为A；
2. 调用operator==(const A&);
3. 调用 A::~A() 来销毁临时文件。

后者是您看到的析构函数调用。

如果您要将以下比较运算符添加到 A，您会看到那些析构函数调用消失了:

  bool operator==(int rhs) { return i_ == rhs; }

或者，如果您像这样调用 remove()，您会看到所有除第一个析构函数调用外的都消失了:

remove( v.begin(), v.end(), A(2) );

最后，如果你要使 A::A(int) explicit，编译器将不允许你调用 remove() 将 2 作为最后一个参数(您必须使用 A(2) 调用它)。