Re: Get lvalue through rvalue

"Igor Tandetnik" <>
Mon, 17 Dec 2007 09:14:49 -0500
"George" <> wrote in message

I do not know how in the following code, rvalue -- return of X(),
could result in a lvalue finally and binded to a non-const reference
input parameter of function f.

Any ideas?

struct X {


void f (X& x) {}

int main()
f (X() = X());

return 0;

There's an implicitly declared compiler-generated assignment operator in
X, that looks like this:

struct X {
  X& operator=(const X&) { return *this; }

So f(X() = X()) is really

f( X().operator=( X() ) );

Now, it is legal to call non-const member functions on temporaries, and
a temporary can be passed as a const X& parameter. Since operator=
returns a non-const reference, the whole thing works similarly to
lvalue_cast we've discussed in previous threads, only less efficiently.
With best wishes,
    Igor Tandetnik

With sufficient thrust, pigs fly just fine. However, this is not
necessarily a good idea. It is hard to be sure where they are going to
land, and it could be dangerous sitting under them as they fly
overhead. -- RFC 1925

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