OpenEuphoria

Re: Assignment Question

• Posted by Pete Lomax <petelomax at blueyonder.co.uk> Feb 26, 2005
• 495 views

On Fri, 25 Feb 2005 09:39:10 -0800, Mario Steele
<guest at RapidEuphoria.com> wrote:

>cause this is really bugging me how the '$' operator messes up everything.

IMHO it doesn't really mess things up, it just changes(for example):

a[idx][idx2]=<expression>

where (bizarrely) <expression> actually changes *a*, to require:

tmp = <expression>
	a[idx][idx2] = tmp

(While I dislike breaking things, I fully accept this as fair, given
we are really only talking about *very* badly written code)

But I didn't reply to this to say that...

>Between the decloration of the variable namespace, the
>assignment operator, and the actual value, there's a type-check done on '1'
>to ensure that it'll fit in 'a', 
NO. Recall the object type. 'a' *is* assigned, and /then/ checked to
be integer-ish. (Basically everything in Eu is an object.)

Try this:

integer  a
a={1,2,3}

You'll get an error/ex.err:
D:\test.exw:2
type_check failure, a is {1,2,3} 

Global & Local Variables

 D:\test.exw:
	a = {1,2,3}

NOTE: a **is** {1,2,3}.

All I am saying is Eu actually assigns, and then checks.
The interpreter *catches* errors, it does not prevent them.

Regards,
Pete