Indirect Addressing Question

[John W. Eaton]

On 13-Nov-2001, Hartmut Henkel <address@hidden> wrote:

| Please excuse the following newbie question:
| 
| With
| 
| a = [1 1 1 1 1]
| q = [1 1 1 2 2]
| 
| the following
| 
| a(q(1:5)) = a(q(1:5)) + 10
| 
| gives
| 
| a = 11 11 1 1 1

In Octave, the RHS is always evalutated first.  So for the above 
expression, you have

  a(q(1:5)) + 10  ==>  a([1, 1, 1, 2, 2]) + 10

                       [1, 1, 1, 1, 1] + 10

                       [11, 11, 11, 11, 11]

then the assignment is performed.  For the above example, we have

  a([1, 1, 1, 2, 2]) = [11, 11, 11, 11, 11]

and this expression says to assign

  a(1) = 11
  a(1) = 11
  a(1) = 11
  a(2) = 11
  a(2) = 11

| But I would expect
| 
| a = 31 21 1 1 1
| 
| which I get by running the loop
| 
| for i = 1 : 5
|   a(q(i)) = a(q(i)) + 10
| endfor
| 
| It seems that such indirect addressing is not allowed. Is there a trick
| to circumvent the for-loop?

I'm not sure of a good way.  Perhaps someone else knows a trick?

BTW, this question has come up before.  Is there some other array
language that works like this?

jwe



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