| Re: LIST "bug": .Remove on an empty list [message #74001 is a reply to message #73999] |
Mon, 13 December 2010 13:48   |
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Paul Van Delst[1]
Messages: 1157
Registered: April 2002
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Senior Member |
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Hello,
I encountered similar behaviour in some code I wrote the other day too, except I did something like
IDL> l=list("a",3.14)
IDL> help, l
L LIST <ID=1 NELEMENTS=2>
IDL> help, l.remove()
<Expression> FLOAT = 3.14000
IDL> help, l.remove()
<Expression> STRING = 'a'
IDL> help, l.remove()
% LIST::REMOVE: Index is out of range.
% Error occurred at: LIST::REMOVE
% $MAIN$
% Execution halted at: $MAIN$
While the error message is not cryptic as in the OP's case, I had assumed that trying to remove something from an empty
list using the remove method would return !null -- similar to how ruby returns "nil" in a similar situation:
irb(main):003:0> l=["a",3.14]
=> ["a", 3.14]
irb(main):004:0> l.pop
=> 3.14
irb(main):005:0> l.pop
=> "a"
irb(main):006:0> l.pop
=> nil
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
Matt Haffner wrote:
> Although this is easy to code around by checking the length before
> calling .Remove, I was surprised this just didn't return silently:
>
> IDL> l=list(1, length=100)
> IDL> help,l
> L LIST <ID=1424588 NELEMENTS=100>
> IDL> l.remove,/all
> IDL> help,l
> L LIST <ID=1424588 NELEMENTS=0>
> IDL> l.remove,/all
> % PTR_FREE: Pointer type required in this context: P.
> % Error occurred at: LIST::REMOVE
> % LIST::REMOVE
> % $MAIN$
> % Execution halted at: $MAIN$
>
> Passing it along to help the diagnosing of cryptic errors ;)
>
> mh
cheers,
paulv
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