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Re: Evaluation of args in pure functions

  • To: mathgroup at
  • Subject: [mg17066] Re: [mg16924] Evaluation of args in pure functions
  • From: "Wolf, Hartmut" <hwolf at>
  • Date: Wed, 14 Apr 1999 02:12:09 -0400
  • Organization: debis Systemhaus
  • References: <>
  • Sender: owner-wri-mathgroup at

Ciao Andrea,

stimulated by your posting (and the replies to it) I did some testing
about the steps of evaluation in different circumstances -- although we
have that scanty description in A.4.1 and A.4.4 for times when we are
not in the mood to scrutinize 2.3, 2.4 once more. Since this test is
funny (and somewhat enlighting, I suppose) I'd like to publish it to the

We try to observe the evaluation sequences in different cases through
added Print statements.

In[1]:= u[x_]:=(Print["hopp"];x)
In[2]:= v[x_]:=(Print["hipp"];x)

In[3]:= u[v[1]]
Out[3]= 1

Ok, v ist evaluated first, then u does it's job at (the result of) v[1]

In[4]:= u[Unevaluated[v[2]]]
Out[4]= 2

First u is called with the (unevaluated) argument v[1], that result is
then evaluated

The operator expr1 @ expr2 essentially is syntactical sugar for
expr1[expr2 ] (except for grouping -> A.2.7)

In[5]:= u at v[3]
Out[5]= 3

In[6]:= u at Unevaluated[v[4]]
Out[6]= 4

So there is no difference.

Coming closer to your question, and testing with pure function:

In[14]:= u[#]& @v[9]
Out[14]= 9

Nothing peculiar, at first the Head of the expression ist evaluated,
i.e. Function  (which we didn't make talking) generates a functional
object (for which we do not have a notation -- it's ephemeral, we can't
see it in the input, we can't see it in the output) without evaluating
its argument (i.e. u[#]). It's only now that we have something that can
be made working, so now  v evaluates, then that functional object moves
the argument (evaluated v[9]) into u, which then executes.

In[16]:= u[#]& @Unevaluated[v[10]]
Out[16]= 10

This is to be contrasted with case 2 above and comes out different! But
we can understand that now: v[10] is passed unevaluated to the
functional object (which moves it into u), but then  v is evaluated
there, before finally u fires.

In[17]:= u[Unevaluated[#]]& @Unevaluated[v[11]]
Out[17]= 11

In[18]:= u[Unevaluated[#]]& @v[12]
Out[18]= 12

So the cases 11 and 12 support our theory for 10. Case 11 holds v
further unevaluated after it is moved into u, executing u first, then
v.  Case 12 is ineffective since v is evaluated before it gets moved
into u.

To get more evidence, you may try to make Function talking:

In[20]:= SetAttributes[talkingFunction, HoldAll]

and do the tests:

In[22]:= talkingFunction[u[#]][v[1009]]

In[23]:= talkingFunction[u[#]][Unevaluated[v[1010]]]

In[24]:= talkingFunction[u[Unevaluated[#]]][Unevaluated[v[1011]]]

In[25]:= talkingFunction[u[Unevaluated[#]]][v[1012]]

So we have to recognize that f[x] and Function[f[#]][x] indeed is not
the same! The expression f[x] is decoded by pattern matching, and the
"arguments" are moved into the body of its definition through named
patterns, resulting in a transformed expression, which then is treated
further. Wheras for Function[f[#]][x] the arguments are identified and
moved into the body by position in the calling sequence, and this is an
_additional_ intermediate step in the calculation process, where further
evaluation of arguments can take place (cases 10, 1010). So we need
Unevaluate twice (cases 11, 1011) to match the case 2.

Finally we can seperate the preparation of the functional object and its
execution into two seperate execution cycles:

In[26]:= foo=talkingFunction[u[#]]

In[27]:= foo[v[2009]]

In[28]:= foo[Unevaluated[v[2010]]]


In[29]:= foouu=talkingFunction[u[Unevaluated[#]]]

In[30]:= foouu[Unevaluated[v[2011]]]

In[31]:= foouu[v[2012]]

(If you want to have my full notebook, just request it by
mailto:hwolf at )

Kind regards
	Hartmut Wolf

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