> > x = y == (x.dm = y.dm) and EQ(x.ob, y.ob)$Lisp
> > which I think is actually not what one wants.
> yes, but this None-business here does not illuminate what is really
> going on.
As far I can tell, EQ means being identical. But (now my guess)
But if you created another string, internally the string is stored in
another memory location.
Does the following session make sense to you? This it what happens. Only
identical elements (i.e., stored in the same memory location) are equal.
Ralf
(1) -> a:String := "A"
(1) "A"
Type: String
(2) -> a1:Any := a
(2) "A"
Type: String
(3) -> a2:Any := a
(3) "A"
Type: String
(4) -> (a1=a2)@Boolean
(4) true
Type: Boolean
(5) -> b:String := "A"
(5) "A"
Type: String
(6) -> (a=b)@Boolean
(6) true
Type: Boolean
(7) -> a3:Any := b
(7) "A"
Type: String
(8) -> (a1=a3)@Boolean
(8) false
Type: Boolean
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