I am a beginner exploring the world of Mathematica. I expected the following code

T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] -> Length[List[b]]}

should return the value 2, rather than 1. Anyone could explain where I was wrong?

  • 4
    $\begingroup$ note that T is a bad choice of variable/function name. I suggest using names starting in lowercase letters, especially when using single letters. This avoids conflicts with built-ins (e.g. N, D...) $\endgroup$ Commented Apr 8, 2013 at 8:26
  • $\begingroup$ Related: (13472), (24860), (26619) $\endgroup$
    – Mr.Wizard
    Commented Jul 15, 2015 at 6:39

2 Answers 2


This is because the code Length[List[b]] is evaluated before the rule is applied. Using RuleDelayed rather than Rule would fix it:

T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] :> Length[{b}]}


The function Trace can be helpful in diagnosing the problem. The documentation says:

Trace returns a set of nested lists. Each individual list corresponds to a single evaluation chain, which contains the sequence of forms found for a particular expression. The list has sublists which give the histories of subsidiary evaluations.

With your input:

  T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] -> Length[List[b]]}

Observe that the first evaluation chain is {Length[{b}],1} which clearly shows the problem. Now compare the Trace when correctly using RuleDelayed:

  T[6, 5, 4, 1, 2, 3] /. {T[a___, 1, b___] :> Length[List[b]]}
  • $\begingroup$ Thank you. This would be very helpful to understand the behavior of Mathematica. $\endgroup$
    – Joonho Kim
    Commented Apr 9, 2013 at 23:23

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