Re: TC++PL(se) section 7.3 Value Return

From:

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <Erik-wikstrom@telia.com>

Newsgroups:

comp.lang.c++

Date:

Sun, 12 Aug 2007 14:06:52 GMT

Message-ID:

<0uEvi.6034$ZA.2834@newsb.telia.net>

On 2007-08-12 15:48, Wayne Shu wrote:

Hi everyone:

    I am reading Bjarne Stroustrup's The C++ Programming
Language(Special Edition).

    In section 7.3, bs wrote "Like the semantics of argument passing,
the semantics of function value return are identical to the semantics
of initialization. A return statement is considered to initialize an
unnamed variable of the returned type."

    Is the *unnamed variable* here always create when we call a
function?

    For instance, there is a function foo,

int foo(int i)
{
    return i*i;
}

when we call it,

int i = foo(5);

Is the above statement mean the following code(not legal in c++)

int tmp = foo(5); // tmp here is the *unnamed variable* bs referred.
int i = tmp;

Even for the build-in types, it is ineffective, why need the *unnamed
variable*?

No, there is something called the named return value optimisation, which
means that instead of creating an temporary and then copying it to the
variable the return value is created directly into the variable. The
following code should demonstrate it:

#include <iostream>

class Foo
{
   int v;
public:
   Foo(int i) : v(i)
   {
     std::cout << "Ctor\n";
   }
   Foo(const Foo& f) : v(f.v)
   {
     std::cout << "Copy Ctor\n";
   }
   Foo& operator=(const Foo& f)
   {
     v = f.v; std::cout << "Assign\n";
   }
};

Foo bar(int i)
{
    return Foo(i);
}

int main()
{
   Foo f = bar(1);
}

If you only get the "Ctor" output the compiler has used the NRVO, if you
also get "Copy Ctor" then it has not.

--
Erik Wikstr??m