I can't define a copy constructor like this:Code:class Expression; Expression func() { Expression x(3); return x; } Expression func2() { return func(); }
So how would I correct the program to silence the error messages saying I need one? I already have a copy constructor looking like this:Code:Expression::Expression (Expression);
Code:Expression::Expression (Expression&);
You'll need to be more specific about your problem, or post some more useful code.
p.s. What the alphabet would look like without q and r.
Let me start off with this question, then I'll ask another if I need to...
Why must copy constructors use references in the arguments?
instead ofCode:Expression::Expression(Expression&);
My compiler tells me that that's an illegal constructor.Code:Expression::Expression(Expression);
Do you know what pass-by-value means? When you don't use a reference, and use this form:
Expression::Expression(Expression);
you are passing by value, so the compiler needs to make a copy of the argument for the function. And guess what? To make a copy, it needs to call the copy constructor. So, your copy constructor calls itself, and once again it needs to make a copy of the argument to the function, so it calls itself again, and so on and so on, resulting in a recursive infinite call to itself.
Last edited by 7stud; 05-11-2003 at 07:31 PM.
I read that the default copy constructor MUST use a reference AND the reference must be const:
Hope that helpsCode:class X{ X(); X(const X&); };
Nothing more to tell about me...
Happy day =)
gustavosserra,
A copy constructor is a function just like any other, and you can make it do what you want:
Code:#include <iostream> using namespace std; class MyClass { public: MyClass(int a) { var = a; } MyClass() { var = 0; } MyClass(MyClass& m) { var = m.var; m.var += 2; } int Get_var(void) { return var; } private: int var; }; int main() { MyClass a(10); cout<<a.Get_var()<<endl; MyClass m = a; cout<<a.Get_var()<<endl; cout<<m.Get_var()<<endl; return 0; }
Last edited by 7stud; 05-11-2003 at 10:36 PM.
it doesnt need to be const but since it doesnt change anything and just copy something to something it is better to make it const to avoid any error that we make.
Yes, I know. But won´t the compiler look for an especific version of the constructor, I mean the version with const. But you are right! I´ve made this code for test:Originally posted by 7stud
gustavosserra,
A copy constructor is a function just like any other, and you can make it do what you want:
The program outputs the both "cout".Code:#include <iostream.h> class MyClass{ public: MyClass(){ cout << "Calling the default constructor\n"; } MyClass(MyClass&){ cout << "Calling the copy constructor\n"; } }; void function(MyClass parameter){} int main(){ MyClass X; function(X); cin.get(); }
Nothing more to tell about me...
Happy day =)
"but since it doesnt change anything..."
Mine does.
gustavosserra,
"But won´t the compiler look for an especific version of the constructor, I mean the version with const."
You tell me. Does the code in my previous post work?
The line:
MyClass m = a;
calls the copy constructor.
Last edited by 7stud; 05-13-2003 at 12:17 AM.
