# Re: rvalue reference questions

From:
David Abrahams <dave@boostpro.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 10 Dec 2008 12:11:59 CST
Message-ID:
<87y6yoc785.fsf@mcbain.luannocracy.com>
on Wed Dec 10 2008, derek-AT-antiquark.com wrote:

Hello,
Something that would aid my understanding greatly would be to learn
about how rvalue references would be converted to a lower level, like C
++.

So, are the following statements correct regarding rvalue references:

The compiler will convert the following code:

double && x = sqrt(123);
cout << x << endl;

to

cout << sqrt(123) << endl;

In other words, 'x' is used as a placeholder for the actual function
sqrt call.

No, not quite. In

cout << sqrt(123) << endl;

the expression "sqrt(123)," if it returns by-value, yields an rvalue.
In

cout << x << endl;

you are dealing with a named entity, 'x'. Named entities are always
lvalues (that's right: even if they're rvalue references).

It's actually possible for C++ code today to distinguish non-const
rvalues from lvalues. For example,

struct catchall
{
template <class T>
catchall(T const&);
};

template <class T>
int f(T&); // called for lvalues and const rvalues

char* f(catchall); // called for non-const rvalues

so

f(sqrt(123))

and

f(x)

would have different results.

Given the function:

void PrintIfGtZero(double && x)
{
if(x > 0) cout << x << endl;
}
...
PrintIfGtZero(sqrt(123));

The compiler will convert the code to:

{
if(sqrt(123))cout << sqrt(123) << endl;
}

No,

Does this mean that sqrt(123) is called twice?

and no. Don't think of rvalue references in terms of code conversion;
it will just get you into trouble. Think of them as regular C++
references with different binding rules and a quirky type-deduction
feature that allows perfect forwarding.

Is is possible that an rvalue reference argument might not be called?

I think if you've understood me up to now, you won't need that question
answered, but just to be clear: arguments are never "called," except in
cases like this one:

int f(int(*g)())
{
return g();
// here-^^
}

HTH,

--
Dave Abrahams
BoostPro Computing
http://www.boostpro.com

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