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Re: Principle of Least Surprise [message #52092]

Tue, 09 January 2007 15:14

Jean H.
Messages: 472
Registered: July 2006

Senior Member

... and
a[3:5].x=bindgen(3)
works perfectly!

It makes sense though... consider this case:
IDL> b=replicate({x:intarr(10)},10)
IDL> b[3].x = indgen(4)

IDL must understand that the specified array is the values of X in b[3].
If what you had written was working, IDL would not know if it is the
values of X or if it is the 1st value of 4 entries of b...

Jean

JD Smith wrote:
>
> Violated again. Consider:
>
> IDL> a=replicate(1b,10)
> IDL> a[3]=bindgen(3)
> IDL> print,a
> 1 1 1 0 1 2 1 1 1 1
>
> but now:
>
> IDL> a=replicate({x:1b},10)
> IDL> a[3].x=bindgen(3)
> % Expression must be a scalar in this context: <BYTE Array[3]>.
> % Execution halted at: $MAIN$
>
> This ought to work.
>
> JD
>

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		Re: Principle of Least Surprise By: Jean H. on Tue, 09 January 2007 15:14
		Re: Principle of Least Surprise By: JD Smith on Tue, 09 January 2007 16:08

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