Re: References to temporaries and function-calls
On Feb 15, 2:51 pm, "Gavin Deane" <deane_ga...@hotmail.com> wrote:
On 15 Feb, 13:16, "Erik Wikstr=F6m" <eri...@student.chalmers.se> wrote:
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,
I'm not sure there is anything else *to* understand.
what I don't understand is
how come this is considered trying to bind a reference to a temporary,
Because the definition of "temporary object" includes the foo object
created in
bar(foo());
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?
Yes to all three, none of which affect the fact that the foo object
created by the statement bar(foo()); is a temporary object.
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)
Yes. An unnamed temporary foo object now exists in the scope of main.
when the
parameter f comes into scope* and thus, from the point of view of
bar() be non-temporary.
The scope of bar isn't where you should be thinking. Assuming you
rewrite bar to take a const foo& so your code compiles, there is no
way (as far as I know) that you can tell *from inside bar* whether the
object referred to by f is a temporary or not in the scope of the
calling function. But from inside bar isn't what's important. What's
important is from inside the function that calls bar (main in this
case).
Inside main, there is a temporary foo object created and you can't
bind non-const references to temporaries. If I add a line to your main
function so you have...
int main() {
bar(foo());
foo& a_reference = foo();
}
...both statements have the same problem. They both try and bind a
temporary to an non-const reference. The fact that in the first case
the non-const reference happens to be a function parameter doesn't
matter. In both cases the object *referred to* is a temporary so the
reference has to be const.
Or put another way, I can't quite see the difference between the
following two:
bar(foo());
and
foo f;
bar(f);
Maybe you can now?
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 just can't see any advantage of the current behaviour over the one I
described nor can I see reason why it should not be possible to
implement (though, admittedly I'm no compiler developer). On the other
hand I can see some usages of allowing the behaviour I described,
among other things a number of algorithms in the standard library
could become more useful.
--
Erik Wikstr=F6m