why does mapping ValueQ produce surprising results?

Question

ClearAll[f1, f2]
f1[x_] := x*x
f2 = #*# &;

This produces the expected results:

ValueQ[f1]  (* False *)
ValueQ[f2]  (* True *)

I find this unexpected:

ValueQ /@ {f1, f2}  (* {False,False} *)

How can I understand the difference?

There is a documentation example about this; use ValueQ /@ Unevaluated[{f1, f2}]. — ilian, Jun 09 '17 at 16:14

Mr.Wizard · Accepted Answer · 2017-06-09T16:51:27.333

2

In ValueQ /@ {f1, f2} the expression {f1, f2} is evaluated before ValueQ is applied, therefore ValueQ never "sees" f2, only its value #*# & which itself does not have a value.

It is critical to understand the standard evaluation order in Mathematica or you shall be chasing many problems or surprises of this nature. Recommended reading:

Operator precedence is also critical unless you exclusively use bracketed notation; see:

when is f@g not the same as f[g]?

edited Jun 09 '17 at 16:51

answered Jun 09 '17 at 16:29

Mr.Wizard

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Thanks. I forgot that ValueQ has the HoldFirst attribute, which should have been obvious. – Alan Jun 09 '17 at 17:58

why does mapping ValueQ produce surprising results?

1 Answers1