Get lvalue through rvalue

[George2]

Hello everyone,


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;
}

thanks in advance,
George

 

[Victor Bazarov]

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.

You're confusing the return type of 'X()' and the full expression
that you can see inside the parentheses where 'f' is called.

Any ideas?

struct X {


};

void f (X& x) {}

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

return 0;
}

The trick here is based on the fact that every temporary of a class
type is a separate object, for which you can call member functions.
The operator= is a member function. What is its usual signature?
Rewrite the expression

f( X() = X() )

in terms of member function calls, and you will hopefully see how
an r-value can be turned into an l-value.

V