lvalue and assignment to temporary object

Consider the following:
int x;
int y;
int z;
(x+y) = z;

For this statement, I get the following error with g++ compiler:
error: non-lvalue in assignment

Suppose I have a class Test and x, y, z are objects of type Test.
Suppose I have
Test Test::operator+(const Test& ref);
Test operator+=(Test lhs, Test rhs);

Given this, the following are accepted by g++:
(x+y) = z;
(x+y) += z;

My question:
Is (x+y) an lvalue ?

I thought, x+y being a temporary object, we cannot take its address
and so not an lvalue. However the compiler accepts it on the left hand
side of the assignment operator. I am unable to understand this.

Kindly clarify

Thanks
V.Subramanian

re: lvalue and assignment to temporary object

On Dec 21, 7:18 am, "subramanian10...@yahoo.com, India"
<subramanian10...@yahoo.comwrote:Correct.
No. But the = here isn't really an operator, either; it's more
a particular syntax of a function call. (Overloaded "operators"
are only partially operators---they have the syntax of
operators, but semantics of a function call.)

Since you can call a member function on an lvalue, the above is
legal.
That would be too simple. x+y is a temporary object (an
rvalue), and *you* cannot take its address. rvalues of class
types do have addresses, however, and the compiler can get at
the address if it needs to, for example to call a member
function.

(Technically, you cannot take its address is only partially
true. You can not apply the built-in operator & on it, since
the built-in operator & requires an lvalue, which it is not. If
the class overloads operator&, however, we're back to the above
rule: the overloaded operator is a function call of a member
function, which is legal.)
It is confusing. There are several different concepts involved
simultaneously, and there is also the fact that rvalue means
something different for class types than for non-class types.

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