复制构造函数理解

Question

我正在通过这个再次研究复制构造函数.

码:

#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;
}

OUTPUT:

Normal constructor allocating ptr
Copy constructor allocating ptr.
Length of line : 10
Freeing memory!
Freeing memory!

我想知道为什么它在调用复制构造函数Line line(10).我认为它应该只调用普通的构造函数.我不是在这里克隆任何对象.请有人解释一下这个.

Answer 1

调用复制构造函数是因为函数参数是按值传递的.这里

void display(Line obj)

该函数定义为Line按值获取参数.这导致正在进行的参数的副本.如果您通过引用传递,则不会调用复制构造函数