Re: Temporary objects, l-values.

"" <>
Tue, 19 Feb 2008 02:36:51 -0800 (PST)
Thanks for the quick reply!

All right, so... now I have lots of nitpicky questions...

"Kai-Uwe Bux" <> wrote in message
news:fpe7i8$trk$ wrote:

Would that also be called, officially,
a "temporary object"?

a) Yes.

Ok. Now if you have this:

  SomeThing x;

There, x itself is not a "temporary object", right, because you
declare it, then use it in another statement -- even though it has a
short life. But if you "shorten" that to this:


Then that SomeThing you pass to function1 *is* a "temporary object".
line is when you ... define the object explicitly in a separate
so it's lifetime is longer than just the expression it was defined in?
Do I
have that right?

b) The code should not compile. The temporary does not bind to non-const
references. See [8.5.3/5].

Thanks for the section reference. I'll read through it tomorrow, it
clear up my confusion.

Nope. You seem to think that something needs to be an l-value to allow for
assignment. That is not true for class types. For class types, the
assignment operator is a member function. You can call non-const member
functions on r-values.

I'm sorry, I don't understand. I'm having a hard time thinking about
When I see people describe what an l-value is, it is usually described
"something that can appear on the left side of an = sign". I take that
mean "something that can be set equal to something else".

So what you are saying is: Because the assignment operator is a
function for class types, a class type does not necessarily have to be
"l-value" to appear on the left side of an = sign, because it's just
calling any other member function. Since I can do
"SomeThing().SomeMemberFunction();", I can do "SomeThing().operator =
(...);" as well, even though it's not an l-value. Right?

The reason that I thought SomeThing() was an l-value is because it
appear on the "left side of an = sign". But, I guess the definition
deeper than that. So... what does it take for a class type value to be
l-value then? If the fact that the = operator is just another member
function like any other means that "SomeThing() = ...;" does not imply
SomeThing() is necessarily an l-value, then it seems like the "=" sign
the "l-value-ness" of a class type are completely unrelated concepts?
other words, if "SomeThing()" can appear on the left side of an =
sign, but
it is not an l-value, then what *is* an l-value (since being able to
be to
the left of an = is not the only requirement)?

Is it more accurate to say an l-value is "something that can be set
equal to
something else", and not worry about the "=" at all... so it's more
the meaning of the statement? What I mean is, if you have a type and
implement an = operator for it that does something else that isn't
assignment, then in that case "=" has nothing to do with l-values.
Similarily, if you have a member function, say, "Assign()" that does
assignment, then an l-value of that type is a value where it's
meaningful to
call "Assign()" on, even though the assignment operator itself is not

I am not disagreeing at all -- I'm just trying to wrap my head around
you typed, because I'm pretty easily confused.

Huh? The reason that 2 = 3 does not make sense is that 2 is const. A() is
not const. Why should it not make sense to set its value?

I'm sorry, "2 = 3" was a really bad example that I used to try to
something that I didn't know the word for. What I meant was something
"in most situations, it's a silly thing to do" (for lack of a better
phrase). Sort of like... assuming the = operator for a type has no
effects, then doing something like "A() = whatever;" doesn't have any
on your program. If you removed it, everything would still be the
That's kind of more what I meant. It was a very bad example.

a) gcc also does not compile the code.

b) Comeau is correct.

It does not compile with GCC... I must have confused myself with
test cases. Argh. Thanks for double-checking that.

Contrary to popular belief, temporaries are not const (unless created
const). However, temporaries do not bind to non-const references.

I did not think temporaries were const. I think I went the extreme in
other direction, trying to assign temporaries to other things. My
logic was
all screwed up. This is what I had thought:

1) L-values are things that appear on the left side of = operators.
2) "(Y(x))" can appear on the left side of = operator.
3) Temporaries are not l-values. Temporaries can not be assigned to
4) Therefore, Y(x) is not a "temporary".
5) Therefore, Y(x) can bind to a non-const reference.

But the flaw in that "proof" (of sorts) is that pretty much every
point is
wrong :( . Because like you said above, the = operator is just
non-const member function for class types, nothing special about it.
temporaries are not necessarily const.


 typeded std::vector< int > int_vector;
 int_vector x;

 x.swap( int_vector() ); // illegal
 swap( x, int_vector() ); // illegal
 int_vector().swap( x ); // legal

In the same way, you can call the assignment operator on a temporary.

If you can call the assignment operator on a temporary, why are you
allowed to bind temporaries to non-const reference parameters? I mean,
like this *seems* entirely reasonable to me:

void function (A &a) {
  // do some stuff to a here

void function2 (void) {

Why is that illegal? You create a new A(), pass a reference to it to
function (so it's not copied when you call function(), it's created
the actual call is made), it's valid inside function(), you can do
stuff to
it, function() returns, statement ends, A() is destroyed. It doesn't
dangerous, unpredictable, indeterminant, or anything. Do you know why
aren't allowed to do that, then (mostly for my own curiosity, I don't
actually have a program that I "need" to do something like that in)?

And since temporaries are not necessarily const, why can they (even
non-const ones) only be bound to const references?

(Usually, there is no point in doing so since all effects of the
will be lost at the end of the full expression.)

*That* is what I was trying to say with the stupid "2=3" example. "No
point". Duh.

What is the underlying problem that you are trying to solve?

It's an off-topic discussion on a Borland newsgroup about exactly
topic. Whether or not you can bind A() to a non-const reference
The source of the confusion is two things:

1) I'm not familiar enough with the standard to know what the real,
behavior is, and
2) The Borland compiler *does*, in fact, accept the above code (with

There was some confusion about which was correct so I asked here to
get a
real answer. The Borland compiler is wrong, of course. It's strange
with the [broken] Borland compiler, the following is accepted, like I

struct X { };

struct Y {
  Y (const X &) { }

void function (Y &) { }

void function2 (void) {
  X x;

But this is not (complaining about binding l-values to non-const

struct Y {

struct X {
  operator Y () { return Y(); }

void function (Y &) { }

void function2 (void) {
  X x;

The difference being that in the first example the conversion uses
constructor, but in the second the conversion uses X's cast operator.
totally weird.

Thanks again for your time!

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