References to temporaries and function-calls

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

struct foo {
int i;
};

int bar(foo& f) {
return f.i++;
}

int main() {
bar(foo());
}

The above code does not compile since you can't bind a reference to a
temporary, you could solve this by using a 'const foo&' parameter
instead but then you have other problems (like trying to change the
value of a const). This much I understand, what I don't understand is
how come this is considered trying to bind a reference to a temporary,
are not all parameters supposed to be evaluated before the function is
executed? And these evaluations should take place in the same scope as
in which the function is called? While the function, including its
parameters, are executed in its own scope?

The way I see things the foo()-part of bar(foo()); should already have
executed (and thus have created a foo-object on the stack) when the
parameter f comes into scope* and thus, from the point of view of
bar() be non-temporary.

Or put another way, I can't quite see the difference between the
following two:
bar(foo());
and
foo f;
bar(f);

Can someone please explain?

* Or is declared, or defined or whatever it is called.

--
Erik Wikström

[Grizlyk]

Erik Wikstrom wrote:

>

>Sorry but no, in both cases the foo-object is temporary, but in one
there's a name and in the other there isn't. I guess I'm kind of
looking for a rationale for this behaviour, and the only thing I can
think of is that if it were allowed you would lose an opportunity to
optimize, namely the ability to create the copied parameter in place
in the stackframe of the function, whereas that would not be possible
if a reference was used.

I do not know wat is rationale. Some compilers can compile that and will
produce warnings if temporary will be created. I have read sometimes ago
something somewhere about, but already forget why.

Sylvester Hesp wrote:

>
In what situations a copy is needed to bind the temporary to a (const)
reference? And if [ foo& a = foo(); ] is not allowed, why is [ foo&
a(foo()); ] allowed? And why does the latter _not_ require a copy ctor?

First [ foo& a = foo(); ] does _not_ require a copy ctor also.

--
Maksim A. Polyanin
http://grizlyk1.narod.ru/cpp_new

"In thi world of fairy tales rolls are liked olso"
/Gnume/

[Victor Bazarov]

Grizlyk wrote:

[..]
Sylvester Hesp wrote:

>>
In what situations a copy is needed to bind the temporary to a
(const) reference? And if [ foo& a = foo(); ] is not allowed, why is
[ foo& a(foo()); ] allowed? And why does the latter _not_ require a
copy ctor?

First [ foo& a = foo(); ] does _not_ require a copy ctor also.

foo& a = foo();

does *not* compile with a fully compliant compiler.

V
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