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I used to thought that calling a function is equal to applying a rule, i.e. applying a rule of f[x_]->sth. is equal to calling a function f with definition f[x_]:=sth.

However, given

disOut[dot[add[y__], z__]] := dot[#, z] & /@ add[y];

dot[add[a, b, c], d, e, f] // disOut (*would produce*)
add[dot[a, d, e, f], dot[b, d, e, f], dot[c, d, e, f]]

while

dot[add[a, b, c], d, e, f] /. dot[add[y__], z__] -> dot[#, z] & /@ add[y](*would be*)
add[dot[a, b, c, d, e, f]]

Why? What's the difference between functions and replacement?

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    – Michael E2
    Commented Jun 7, 2015 at 14:03

1 Answer 1

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Some parentheses and RuleDelayed instead of a Rule:

dot[add[a, b, c], d, e, f] /. dot[add[y__], z__] :> (dot[#, z] & /@ add[y])
(*  add[dot[a, d, e, f], dot[b, d, e, f], dot[c, d, e, f]]  *)

The :> prevents dot[#, z] & /@ add[y] from evaluating until after y and z have been replaced by the expressions they matched.

DownValues shows the rules that a function evaluation uses:

DownValues[disOut]
(*  {HoldPattern[disOut[dot[add[y__], z__]]] :> (dot[#1, z] &) /@ add[y]}  *)

Note that the parentheses may be put in another place, too. The reason you need the parentheses has to do with the precedence of the operators & and Rule/RuleDelayed.

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  • $\begingroup$ Thanks a lot! So functions are (delayed) rules after all? $\endgroup$
    – user372021841
    Commented Jun 7, 2015 at 14:26
  • $\begingroup$ @user372021841 Yes, but the order in which functions and their arguments are evaluated can be influenced by attributes. See tutorial/Evaluation, if you haven't already seen it. $\endgroup$
    – Michael E2
    Commented Jun 7, 2015 at 15:11
  • $\begingroup$ Not quite. Do a search for functions vs. Patterns. $\endgroup$
    – LLlAMnYP
    Commented Jun 7, 2015 at 16:11
  • $\begingroup$ @LLlAMnYP I believe that is mentioned in the tutorial I linked to, but thanks for a hint to the link for further discussion on the site. The issues are rather broad, technical, and complicated to address, I felt. But when a function like the OP's is evaluated, it is a rule that is being applied, isn't it? $\endgroup$
    – Michael E2
    Commented Jun 7, 2015 at 17:06
  • $\begingroup$ Yes, the OP gives a replacement rule, function or not on the right hand side. I think that the SetDelayed way of giving DownValues is in a way also defining a replacement rule. I can't say much more, as I'm not that fluent on the internals of Mathematica. $\endgroup$
    – LLlAMnYP
    Commented Jun 7, 2015 at 18:48

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