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Re: Count[{Unevaluated[a]},Unevaluated[a]]

  • To: mathgroup at yoda.physics.unc.edu
  • Subject: Re: Count[{Unevaluated[a]},Unevaluated[a]]
  • From: John Lee <lee at math.washington.edu>
  • Date: Wed, 2 Dec 92 10:32:47 -0800

>  the above gives - a little bit surprisingly for me - 0.
>  With a third argument: Count[{Unevaluated[a]},Unevaluated[a],{-1}]
>  the result is 1 (as expected). Using On[] one sees that 
>  Count "evaluates" its arguments. Literal instead of Unevaluated gives
>  a similar result.
>  Why does Count do this?

Try Count[Unevaluated[{a}],Unevaluated[a]].  In your form, {Unevaluated[a]}
passes a unevaluated to List, which results in [a], which is then passed to
Count.  

Jack Lee
Dept. of Mathematics
University of Washington
Seattle, WA 





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