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Re: [HTCondor-users] Comparing evaluation result with "classad.Value.Error" throws error

Hi Ceyhun,

Indeed, this is a subtle one!  Excellent question.

Short answer: You want "value is classad.Value.Error".

Long answer:

The _expression_ "value == classad.Value.Error" results in a classad.ExprTree object which has the representation "error == error".  The "if" clause coerces the _expression_ to a boolean.  Thus, "error == error" is evaluated to the classad Value object "error" (as comparing anything to an error is still an error; "1 == error" evaluates to an error as well!)

Now, what's the boolean value of "error"?  Before the 8.9 series, boost converted error to the integer value "2" (sadly, this is the C++ enumerated value) and hence "error" returned "true" because bool(2) is True in python.  This was tricky because an _expression_ like "1 == classad.Value.Error" would also return true as well as "classad.Value.Undefined == classad.Value.Error".  Whoops!

In 8.9, converting an "error" object to a boolean raises an exception to avoid this mistake.

Instead, "value is classad.Value.Error" returns whether value and classad.Value.Error are the same object instead of trying to evaluate a ClassAd _expression_.  Hence, your code can be simplified to:

value = ad.eval(key)
if value is classad.Value.Error:

(It's unfortunate but these are the sort of tricky edge cases that appear when trying to model the subtly-different boolean logic of ClassAds in a pythonic manner.)


On May 8, 2021, at 2:27 PM, Ceyhun Uzunoglu <ceyhun.uzunoglu@xxxxxxx> wrote:

Dear htcondor community,
Iâm upgrading my codebase to htcondor v9.0.0 from v8.7.9.
As it is mentioned in this comment[1], comparison of âclassad.Value.Errorâ throws âclassad.ClassAdEvaluationError: Unable to evaluate _expression_.â exception.
What do you suggest to use instead of line-3. Iâm trying to check evaluation result is not a classad.Value.Error.
1. value = ad.eval(key)
2. if isinstance(value, classad.Value):
3.    if value == classad.Value.Error:
Kind regards,
Ceyhun Uzunoglu
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