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Re: The uses of Condition (/;)

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  • Subject: [mg82956] Re: [mg82926] The uses of Condition (/;)
  • From: "Szabolcs HorvÃt" <szhorvat at>
  • Date: Mon, 5 Nov 2007 04:56:24 -0500 (EST)
  • References: <>

On 11/3/07, Ingolf Dahl <ingolf.dahl at> wrote:
> I think the real example in case C should be
> h1[x_] := Module[{}, (Print["def"]; x) /; (Print["cond"]; x > 5)]
> because your h[x_] expression is interpreted as
> h[x_] := Module[{}, Print["def"]; (x /; (Print["cond"]; x > 5))]
> which does not correspond to the case C. It instead corresponds to
> D. pattern := Module[{}, expr1; (definition /; condition)]
> This leads us to an answer to your question. We define a new pattern, a
> modification of pattern B:
> E. pattern := (expr1; (definition /; condition))
> with the example
> g1[x_] := (Print["expr1"]; ((Print["def"]; x) /; (Print["cond"]; x > 5)))
> In[8]:= g1[0]
> During evaluation of In[8]:= expr1
> During evaluation of In[8]:= cond
> Out[8]= g1[0]
> This function behaves similar to the D function, and we see that Module is
> just a wrapper. Module here is useful if we want to share local variables
> between the definition and the expression, so that we can transfer sub
> results and do not have to repeat calculations.
> I do not see how to modify A to get the same kind of evaluation as with the
> case E.
> See also the MathGroup discussion about precedence of operators earlier this
> week.

Thanks for the reply (and for pointing out that /; has a higher
precedence than ;)!

So in fact cases B and C are sub-cases of the same usage, and Module
has no importance.  All that matters is that Condition[] be the
expression that would be returned by the function.  (Well, almost:
Condition[] must be returned by a restricted set of built-ins, like
CompoundExpression or Module.  f[x_] := Identity[x /; x > 5] does not
work.  But this is understandable, since in Mathematica the "return
value" is not a well defined concept.)


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