Re: lvalue required as increment operand -- why does the Standard requires this for fundamental types only?

From:
SG <s.gesemann@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 4 Jul 2011 04:07:22 -0700 (PDT)
Message-ID:
<03702289-d21e-4ca9-9c05-f6428ce66af1@d22g2000yqn.googlegroups.com>
On 3 Jul., 20:51, Pavel wrote:

[...]

  template <class C> void foo(C &c) {
     assert(!c.empty());
     typename C::iterator lastElemIter = --c.end();
     ... // use lastElemIter here
  }

would give me an iterator to the last element. It works if the iterator's
implementation happen to be a class


Not necessarily. operator-- could be implemented as a free function
taking an lvalue reference parameter. And it would be ok to do so
because the ITERATOR CONCEPT only requires ++ and -- (for
bidirectional iterators) to work on LVALUES. We have interfaces for a
reason. Don't rely on specific implementations!

On 3 Jul., 21:12, Pavel wrote:

SG wrote:

   template<class Iter>
   inline Iter next(Iter it) { std::advance(it,1); return it; }


This is essentially what I need semantically. Performance-wise, however, =

I want

to avoid std::advance, because it may be and often is less efficient for =

many

iterators (e.g. list or map iterators -- which are usually class object a=

nd for

which ++ would work).


These are bidirectional iterators. For such iterators std::advance
maps to something like this via tag dispatching:

   template<class Iter>
   inline void advance_(
           Iter & it,
           typename iterator_traits<Iter>::difference_type d,
           bidirectional_iterator_tag)
   {
       if (d<0) {
         while (d++ < 0) --it;
       } else {
         while (d-- > 0) ++it;
       }
   }

Compared to ++it, advance(it,1) only requires two additional tests --
that is, unless the compiler inlines this function and applies some
dead branch elimination because the value d is known to be 1 at
compile time during inlining.

Of course, why write advance(it,1) if you can write ++it directly? I
don't know. That's just how the upcoming standard describes the
behaviour of std::next. I guess implementers will simply write this

   template<class Iter>
   inline Iter next(Iter x) { ++x; return x; }

considering the "as-if rule".

And just for the fun of it, let's ACTUALLY test how G++ optimizes
advance with a constant second parameter of value 1 for list
iterators:

  #include <iterator>
  #include <list>

  typedef std::list<int>::iterator iter;

  iter next(iter it) {
    std::advance(it,1);
    return it;
  }

--[ g++ -O3 advance.cpp && objdump -Cd advance.o ]-->

  00000000 <next(std::_List_iterator<int>)>:
   0: 55 push %ebp
   1: 89 e5 mov %esp,%ebp
   3: 8b 45 08 mov 0x8(%ebp),%eax
   6: 8b 00 mov (%eax),%eax
   8: c9 leave
   9: c3 ret
   a: 90 nop
   b: 90 nop

I see no loop, no tests of any kind. In fact, when I replace
advance(it,1) with ++it, I see EXACTLY the same assembly code.
Hmm.... :-D

Cheers!
SG

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