c++ - What is copy elision and how does it optimize the copy-and-swap idiom?

Question

I was reading Copy and Swap.

I tried reading some links on Copy Elision but could not figure out properly what it meant. Can somebody please explain what this optimization is, and especially what is mean by the following text

This is not just a matter of convenience but in fact an optimization. If the parameter (s) binds to a lvalue (another non-const object), a copy of the object is made automatically while creating the parameter (s). However, when s binds to a rvalue (temporary object, literal), the copy is typically elided, which saves a call to a copy constructor and a destructor. In the earlier version of the assignment operator where the parameter is accepted as const reference, copy elision does not happen when the reference binds to a rvalue. This results into an additional object being created and destroyed.

Answer 1

The copy constructor exists to make copies. In theory when you write a line like:

CLASS c(foo());

The compiler would have to call the copy constructor to copy the return of foo() into c.

Copy elision is a technique to skip calling the copy constructor so as not to pay for the overhead.

For example, the compiler can arrange that foo() will directly construct its return value into c.

Here's another example. Let's say you have a function:

void doit(CLASS c);

If you call it with an actual argument, the compiler has to invoke the copy constructor so that the original parameter cannot be modified:

CLASS c1;
doit(c1);

But now consider a different example, let's say you call your function like this:

doit(c1 + c1);

operator+ is going to have to create a temporary object (an rvalue). Instead of invoking the copy constructor before calling doit(), the compiler can pass the temporary that was created by operator+ and pass that to doit() instead.