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Re: same mem problem

From: John W. Eaton
Subject: Re: same mem problem
Date: Tue, 8 Feb 2000 22:19:19 -0600 (CST)

On  0-Feb-, Erich Schneider <address@hidden> wrote:

| Try this code and watch the memory usage:
| clear;
| f5hs3=rand(570,720)*5000;
| f5ls3=rand(570,720)*5000+1;
| C=-636.73*(f5hs3-f5ls3)./f5ls3-0.079098*f5hs3+332.15;
| clear f5ls3;
| [m,n]=size(f5hs3);
| C_temp1=C(2:2:m,2:2:n);
| clear C;
| C_temp=C_temp1(:);
| clear C_temp1;
| average=mean(C_temp(find(f5hs3(2:2:m,2:2:n)<3400)))
| clear C_temp;
| You could even subdivide the calculation for C into something like:
| c1 = f5hs3-f5ls3;
| c2 = -636.73.*c1./f5ls3;
| clear c1;
| C = c2+332.15-0.079098.*f5hs3;
| clear c2;
| I'm using 2.0.13, too, but couldn't reproduce your error when running
| your script repeatedly. It seems to me that for the calculation of C
| octave is storing 2 additional temporary matrices.

If you mean the calculation of


then I think the best Octave can do is

  t1 = (f5hs3-f5ls3)
  t2 = -636.73*t1
  t1 = t2 ./ f5ls3
  t2 = 0.079098*f5hs3
  t3 = t1 - t2
  t1 = t3 + 332.15;
  C = t1 # just some reference counting magic; elements are not acutally copied

so yes, it takes some temporary storage.  Octave doesn't try to
optimize this operation into one loop.  If you can do that, please
submit a patch!  AFAIK, it would not be trivial...

| I saw the following behaviour of octave 2.0.13 and .14:
| a=ones(8000000,1);
| a=ones(8000000,1);
| memory exhausted bla ...
| b=ones(4000000,1);
| c=ones(4000000,1);
| If memory is exhausted then why can I execute the steps for b and c?

The `memory exhausted' message means
that there wasn't enough memory available to satisfy the request, not
that there is no more memory available.  So it is not surprising that
the allocation for `b' succeeded after you tried to allocate two 
8000000x1 matrices (two, because Octave doesn't delete the contents of
the left-hand side of the assignment until the right-hand side has
been evaluated; this is a feature, not a bug).  It's hard to say why
the allocation for `c' also succeeded without knowing more about what
might have been happening on your system at the time.  Perhaps some
memory was freed by some other process and that allowed your
allocation to succeed?


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