1. Maybe slightly interesting, perhaps
No, it doesn't matter, but someone may be mildly curious (Rob?):
type integer(integer i)
return 1
end type
The interpreter goes into an infinite loop. I think I can guess why.
I was trying to redefine integer to be 0..33, to track down a bug,
I had about 15 variables, and once one went, domino effect, (long
before it actually crashed), but a global search/replace did the trick
so no worries.
Regards,
Pete
http://palacebuilders.pwp.blueyonder.co.uk/euphoria.html
2. Re: Maybe slightly interesting, perhaps
Pete Lomax wrote:
> No, it doesn't matter, but someone may be mildly curious (Rob?):
>
> type integer(integer i)
> return 1
> end type
>
> The interpreter goes into an infinite loop. I think I can guess why.
The redefinition of integer occurs before it can be used for the
parameter to the redefined type, so yes, the new type calls itself.
The standard Euphoria idiom is (used to be?) this kind of construct:
type eu_integer(integer i)
return 1
end type
without warning
type integer(eu_integer i)
return 1
end type
with warning
The above works because Euphoria doesn't do mutual recursion (by default).
<soapbox>
Then again, I favour the following method of type definintion as it
makes them less likely to crash the interpreter with a type-check error
(Oh, the irony...):
type eu_integer(object i) if integer(i) then
return 1
end if return 0 end type
without warning
type integer(eu_integer i) if eu_integer(i) then
return 1
end if return 0 end type
with warning
</soapbox>
Carl
--
[ Carl R White == aka () = The Domain of Cyrek = ]
[ Cyrek the Illogical /\ www.cyreksoft.yorks.com ]
3. Re: Maybe slightly interesting, perhaps
Carl W. wrote:
> without warning
> type integer(eu_integer i) if eu_integer(i) then
Oops. That should read:
without warning
type integer(object i) if eu_integer(i) then
Sorry about that.
Carl
--
[ Carl R White == aka () = The Domain of Cyrek = ]
[ Cyrek the Illogical /\ www.cyreksoft.yorks.com ]
