| Re: LIST "bug": .Remove on an empty list [message #73999 is a reply to message #73994] |
Mon, 13 December 2010 19:18   |
penteado
Messages: 866
Registered: February 2018
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Senior Member
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On Dec 13, 7:48 pm, Paul van Delst <paul.vande...@noaa.gov> wrote:
> But I got the "LIST::REMOVE: Index is out of range" error instead. I'm still trying to figure out how to best handle it
> (it does make using lists a little bit more complicated). The HASH remove method works as I expected:
>
> IDL> h=hash("a", 3.14)
> IDL> help, h
> H HASH <ID=1 NELEMENTS=1>
> IDL> help, h.remove()
> <Expression> FLOAT = 3.14000
> IDL> help, h.remove()
> <Expression> UNDEFINED = !NULL
Besides the bad error message for list::remove(), I would say that the
bug is in hash::remove(). Unless I am missing something, it should
throw an error on an empty hash. Returning !null with no errors makes
the empty list/hash look like it has elements (which happen to have !
null as a value).
For instance,
IDL> l=list(length=1)
IDL> help,l.remove()
<Expression> UNDEFINED = !NULL
IDL> help,l.remove()
% LIST::REMOVE: Index is out of range.
% Error occurred at: LIST::REMOVE
% $MAIN$
% Execution halted at: $MAIN$
If the second remove() returned !null without errors, it would have
looked like the first call, which was properly returning (and
removing) the !null which was the list element. List is doing things
properly, it is hash that is confusing:
IDL> h=hash('a')
IDL> help,h
H HASH <ID=1 NELEMENTS=1>
IDL> help,h.remove()
<Expression> UNDEFINED = !NULL
IDL> help,h
H HASH <ID=1 NELEMENTS=0>
IDL> help,h.remove()
<Expression> UNDEFINED = !NULL
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