爲什麼這個代碼在傳遞的對象不是Line類型並且沒有等於操作符/顯式調用時調用拷貝構造函數。線A和線A()之間有區別。
我從網上很多教程,它應該是線路type.I的閱讀是一個新手,C++請幫忙使用不同的參數在C++中拷貝構造函數
#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;
}
_「Line A;'和'Line A();'?」_ [Yes]是否有區別(https://en.wikipedia.org/wiki/Most_vexing_parse)。 – nwp