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Re: HoldFirst, Unevaluated
*To*: mathgroup at smc.vnet.net
*Subject*: [mg109627] Re: HoldFirst, Unevaluated
*From*: Albert Retey <awnl at gmx-topmail.de>
*Date*: Sat, 8 May 2010 07:09:16 -0400 (EDT)
*References*: <hs0pp2$dsn$1@smc.vnet.net>
Am 07.05.2010 12:24, schrieb Fred Klingener:
> Here, the second execution of f fails because variable 'a' evaluates
> to Pi and can't be reassigned:
>
> ClearAll[f, a, b, c]
> f[x_, y_, z_] := x = Pi; 3 y + 2 z
> f[a, b, c]
> f[a, b, c]
>
> I have come to understand that the canonical approach to control this
> is to assign the HoldFirst attribute to f:
>
> ClearAll[f, a, b, c]
> SetAttributes[f, HoldFirst]
> f[x_, y_, z_] := x = Pi; 3 y + 2 z
> f[a, b, c]
> f[a, b, c]
> a
>
> and, indeed, this works as intended.
>
> Plan B, using an explicit Unevaluated also works:
>
> ClearAll[f, a, b, c]
> f[x_, y_, z_] := x = Pi; 3 y + 2 z
> f[Unevaluated[a], b, c]
> f[Unevaluated[a], b, c]
> a
>
> An alternate function formulation fails, evidently in the same way as
> the first example:
>
> ClearAll[f, a, b, c]
> SetAttributes[f, HoldFirst]
> f = (#1 = Pi; 3 #2 + 2 #3) &
> f[a, b, c]
> f[a, b, c]
> a
You probably want this:
ClearAll[f, a, b, c]
f = Function[Null,#1 = Pi; 3 #2 + 2 #3,HoldFirst]
>
> ClearAll[f, a, b, c]
> f = (#1 = Pi; 3 #2 + 2 #3) &
> f[Unevaluated[a], b, c]
> f[Unevaluated[a], b, c]
> a
>
> does.
>
> What am I missing?
Attributes of Symbols affect only the downvalues you have defined for
them, in the case that did not work, you did assign an ownvalue, which
happens to be a function. Thus the HoldFirt did not have an effect,
since the Function still did evaluate its arguments. You can set
attributes to Functions as shown above...
hth,
albert
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