[LC++]can't seem to use reference

[Carlo Wood]

On Mon, Feb 18, 2002 at 11:03:06AM +1030, Mark Phillips wrote:
> Carlo Wood wrote:
> > 
> > On Tue, Feb 12, 2002 at 01:27:31PM +1030, Mark Phillips wrote:
> > > largeStructTy const& nextLs=largeStructList[0];
> > > for (int i=0; i<N; ++i) {
> > >   largeStructTy const& ls=nextLs;
> > >   nextLs=largeStructList[i+1];
> > >   // do stuff involving the reading of ls and nextLs
> > >   // ...
> > > }
> > 
> > int next=0;
> > for (int i=0; i<N; ++i) {
> >   largeStructTy const& ls=largeStructList[next];
> >   largeStructTy const& nextLs=largeStructList[(next=i+1)];
> >   // do stuff involving the reading of ls and nextLs
> >   // ...
> > }
> 
> Thanks for this solution.  My only concern with it is the
> possibility of inefficiency.  My solution (if it were legal C++)
> would only require one large structure to be read from memory
> each iteration (the rest is simply pointer swapping).  Your
> solution requires two large structures to be read each iteration.
> Or would the compiler optimize this away?

My 'solution' doesn't copy anything at all, there are only
reference initializations.

TYPE& foo = initializer;

is equivalent with

TYPE& foo(initializer);

and functionally equivalent with

TYPE* const foo_ptr = &initializer;
// use (*foo_ptr)

Actually, I hadn't noticed that you copied data from 'i+1' to 'i';
So I should have typed:

for (int i = 0; i < N; ++i)
{
  largeStructTy const& ls(largeStructList[i]);
  largeStructTy const& nextLs(largeStructList[i]);
  nextLs = largeStructList[i + 1];
  // do stuff involving the reading of ls and nextLs
  // ...
}

which is equivalent with your pseudo code.
Note that there is no difference at all between ls and nextLs.

I can assume that what you really want is this:
Copy element i+1 to i, keep a reference to both, where
reference ls corresponds to i and nextLs to i+1.

That would be:

for (int i = 0; i < N; ++i)
{
  largeStructTy const& ls(largeStructList[i]);
  largeStructTy const& nextLs(largeStructList[i+1]);
  ls = nextLs;
  // do stuff involving the reading of ls and nextLs
  // ...
}

> Ie, why the use of "next"?

I applied a 'code transformation' from "lsNext as variable" to
"lsNext as reference with variable index 'next'".  Since 'next'
and 'i' are related, you can eliminate 'next' again ofcourse.
I thought that not 'optimizing' the code would make it more
clear what I had done.

-- 
Carlo Wood <carlo at alinoe.com>