| Re: Optimising A = B+C? [message #6345 is a reply to message #6257] |
Sun, 02 June 1996 00:00  |
davis
Messages: 15
Registered: March 1995
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Junior Member |
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On Fri, 31 May 1996 02:24:03 GMT, Karl Glazebrook <kgb@aaoepp.aao.gov.au>
wrote:
: Does anyone know how IDL optimises A=B+C where
: A,B and C are arrays?
:
: I did a test a while ago and it was several times faster
: than C code along the lines of:
:
: i=n;
: while (i--)
: *a-- = *b-- + *c--
Although I have no proof, I believe that the above can be coded to run
faster, e.g.,
for (i = 0; i < n; i++) a[i] = b[i] + c[i];
Also, if IDL arrays are allocated with an even number of elements (with the
last one padded for odd sized arrays), this is faster:
for (i = 0; i < n; i++)
{
a[i] = b[i] + c[i];
i++;
a[i] = b[i] + c[i];
}
:
: (This was on a 2048x2048 array and everything fiited into
: physical memory.)
:
It also depends upon how your two-dimensional array was implemented. One
common C approach is to use, e.g.,
double **a;
unsigned int i;
a = malloc (2048 * sizeof (double *));
for (i = 0; i < 2048; i++)
a[i] = malloc (2048 * sizeof(double));
Now consider adding such arrays together. One might code this as
for (i = 0; i < 2048; i++)
{
for (j = 0; j < 2048; j++)
{
a[i][j] = b[i][j] + c[i][j];
}
}
I am not sure how many compilers will optimize this. For that reason, I
never code loops like the above. Instead, I always do
for (i = 0; i < 2048; i++)
{
double *ai, *bi, *ci;
ai = a[i]; bi = b[i]; ci = c[i];
for (j = 0; j < 2048; j++)
{
ai[j] = bi[j] + ci[j];
}
}
Finally, if you simply create such arrays via:
double a[2048][2048]; /* this is just a 2048*2048 block of doubles */
then you might be tempted to perform the addition as
for (i = 0; i < 2048; i++)
{
for (j = 0; j < 2048; j++)
{
a[i][j] = b[i][j] + c[i][j];
}
}
The problem with this is that it is not very friendly to your CPU cache
because the loop over j does not deal with neighboring values in the array.
As a result, it is best to calculate the sum for these types of arrays as:
double *aa, *bb, *cc;
unsigned int i;
aa = (double *) a;
bb = (double *) b;
cc = (double *) c;
for (i = 0; i < 2048*2048; i++)
{
aa[i] = bb[i] + cc[i];
}
--
John E. Davis Center for Space Research/AXAF Science Center
617-258-8119 MIT 37-662c, Cambridge, MA 02139
http://space.mit.edu/~davis
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