copy-constructor related question (native c++) [duplicate]

Question

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

i know that a copy constructor must have a reference as a parameter, to avoid an 'infinite number of calls' to itself. my question is - why exactly that happens, what is the logic behind it?

CExample(const CExample& temp)
{
   length = temp.length;
}

Answer 1

assume your argument to the copy C'tor was passed by value, the first thing the C'tor would have done, was copying the argument [that's what every function, including constructors do with by-value arguments]. in order to do so, it would have to invoke the C'tor again, from the original to the local variable... [and over and over again...] which will eventually cause an infinite loop.

Answer 2

Copy constructors are called on some occasions in C++. One of them is when you have a function like

void f(CExample obj)
{
    // ...
}

In this case, when you make a call

CExample x;
f( x );

CExample::CExample gets called to construct obj from x.

In case you have the following signature

void f(CExample &obj)
{
     // ...
}

(note that obj is now passed by reference), the copy constructor CExample::CExample does not get called.

In the case your constructor accepts the object to be copied by value (as with the function f in the first example), compiler will have to call the copy constructor first in order to create a copy (as with the function f, again), but... oops, we have to call the copy constructor in order to call the copy constructor. This sounds bad, doesn't it?