Elimination of one occurrence of variable result in

  result:=[]
  every ti:=!X.member_of do
    put(result, something)

is possible (thanks to Steve Hunter) with every put(result:=[],something(!X.member_of))

That surprised me, because I thought that every resumes whole expression. Is it possible to do it more generally?

result1:=[]
result2:=[]
every ti:=!X.member_of do
    { put(result1, something1(ti))
      put(result2, something2(ti))
    }



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Kazimir Majorinc, Zagreb, Croatia

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