copy constructor in c++

Question

Possible Duplicate:
Why should the copy constructor accept its parameter by reference in C++?

why do we have have a reference always in a copy constructor?

if there is a class:

class Base{};

its copy constructor is:

Base::Base(const Base& B);

why is always a reference as the argument?

score 14 · Answer 1 · edited Aug 09 '19 at 13:07

14

Had it not been a reference, the compiler would need to invoke the copy constructor in order to pass the parameter to the copy constructor.

That would result in a
_{(source: prodeveloper.org)}
.

edited Aug 09 '19 at 13:07

Glorfindel

21,988
13
81
109

answered Mar 15 '11 at 13:00

SLaks

868,454
176
1,908
1,964

very bad.. someone is getting so many upvotes :P – Nawaz Mar 15 '11 at 13:06

score 0 · Answer 2 · answered Mar 15 '11 at 13:01

0

If you tried to pass the parameter by value the compiler would have to copy the parameter and that would again invoke the copy constructor. This would lead to endless recursion and stack overflow.

answered Mar 15 '11 at 13:01

sharptooth

167,383
100
513
979

score 0 · Answer 3 · answered Mar 15 '11 at 13:04

copy-constructor is called when there is a need of making a copy of your object.

Now think, if the copy-constructor doesn't take reference of the object, that means you've to pass it by value that in turn would require making a copy of your original object. That means, it'll again call copy-constructor, to make a copy of the object. In this way, it'll keep on calling copy-constructor.

copy constructor in c++

3 Answers3