Re: a=a(*,*,[4,1,2,3,0]) efficiency

[davis(at)space.mit.edu (John E. Davis)]

On 15 Jul 1998 01:30:30 +0200, David Kastrup <dak@mailhost.neuroinformatik.ruhr-uni-bochum.de>
wrote:
>temporary in the first place.  How about
>
>a = (temporary(a))[*,*,[4,1,2,3,0]]

I have no knowledge of the internals of IDL, but I do not think that
the use of `temporary' will help.  I am guessing that `temporary'
simply does the following:

     1.  Push value of `a' onto the stack.  This results in the
	 reference count to array attached to `a' being increased by 1.

     2.  Free `a' and undefine the variable.  This has the effect of
	 decrementing the reference count of array attached to `a' by 1.

The net result is that the ownership of the array attached to `a' will
have changed from `a' to the stack.  Now consider:

   a = a[*,*,[4,1,2,3,0]]

This will probably do the following:

    1.   Push value of `a' onto stack.  Reference count of array
	 increased by 1.
	 
    2.   Retrieve array from stack.

    3.   Create a new array that is a copy of the array on the stack
	 but with elements interchanged.  Push result onto stack with
	 a reference count of 1.
	 
    4.   Free array popped from stack.  This reduces the reference
	 count of array attached to `a' by 1.

    5.   Assign the value of array on stack to `a'.  First free the
	 array attached to `a', reducing the reference count by 1.
	 
    6.	 Then remove the new array from the stack and assign it to
	 `a'.  The reference count of this array is still 1.

In both cases, at some instant, the original array and its
``interchanged'' copy will both exist.  All `temporary' does is move
step 5 to between steps 1 and 2.

I imagine that `temporary' is really only useful in more complex
expressions, e.g., consider

     a = (a + b) + c

which consists of 3 arrays `a', `b', and `c'.  During the evaluation
of the RHS of this statement, 2 extra arrays will be created: (a+b)
and the result (a+b)+c.  Thus at some point, 5 arrays will exist.
Just prior to the assignment to `a', the temporary arrat (a+b) will be
freed.  Now consider:

     a = (temporary(a) + b) + c

After the evaluation of (temporary(a)+b), only 3 arrays will exist:
(a+b), b, and c.  Then when (a+b) is added to `c', another array will
be created raising the total number needed to 4.

Again, this is pure speculation and I may be totally wrong.  But I
cannot thing of another way to implement this.

--John