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  C++ 拷贝构造函数

  • 拷贝构造函数

    拷贝构造函数是通过用相同的类，它已先前创建的一个目的初始化它创建了一个对象的构造。复制构造函数用于-
    • 从另一个相同类型的对象初始化一个对象。
    • 复制对象以将其作为参数传递给函数。
    • 复制对象以从函数返回它。
    如果没有在类中定义副本构造函数，则编译器本身将定义一个副本构造函数。如果该类具有指针变量并具有一些动态内存分配，则必须具有副本构造函数。复制构造函数的最常见形式如下所示： 
    
classname (const classname &obj) {
   // body of constructor
}
    此处，obj是对用于初始化另一个对象的对象的引用。 
    
#include <iostream>

using namespace std;

class Line {

   public:
      int getLength( void );
      Line( int len );             // simple constructor
      Line( const Line &obj);  // copy constructor
      ~Line();                     // destructor

   private:
      int *ptr;
};

// Member functions definitions including constructor
Line::Line(int len) {
   cout << "Normal constructor allocating ptr" << endl;
   
   // allocate memory for the pointer;
   ptr = new int;
   *ptr = len;
}

Line::Line(const Line &obj) {
   cout << "Copy constructor allocating ptr." << endl;
   ptr = new int;
   *ptr = *obj.ptr; // copy the value
}

Line::~Line(void) {
   cout << "Freeing memory!" << endl;
   delete ptr;
}

int Line::getLength( void ) {
   return *ptr;
}

void display(Line obj) {
   cout << "Length of line : " << obj.getLength() << endl;
}

// Main function for the program
int main() {
   Line line(10);

   display(line);

   return 0;
}
    尝试一下
    编译并执行上述代码后，将产生以下结果- 
    
Normal constructor allocating ptr
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Freeing memory!
    让我们看一下相同的示例，但是稍作改动即可使用相同类型的现有对象创建另一个对象- 
    
#include <iostream>

using namespace std;

class Line {
   public:
      int getLength( void );
      Line( int len );             // simple constructor
      Line( const Line &obj);  // copy constructor
      ~Line();                     // destructor

   private:
      int *ptr;
};

// Member functions definitions including constructor
Line::Line(int len) {
   cout << "Normal constructor allocating ptr" << endl;
   
   // allocate memory for the pointer;
   ptr = new int;
   *ptr = len;
}

Line::Line(const Line &obj) {
   cout << "Copy constructor allocating ptr." << endl;
   ptr = new int;
   *ptr = *obj.ptr; // copy the value
}

Line::~Line(void) {
   cout << "Freeing memory!" << endl;
   delete ptr;
}

int Line::getLength( void ) {
   return *ptr;
}

void display(Line obj) {
   cout << "Length of line : " << obj.getLength() << endl;
}

// Main function for the program
int main() {

   Line line1(10);

   Line line2 = line1; // This also calls copy constructor

   display(line1);
   display(line2);

   return 0;
}
    尝试一下
    编译并执行上述代码后，将产生以下结果- 
    
Normal constructor allocating ptr
Copy constructor allocating ptr.
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Freeing memory!
Freeing memory!
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