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Re: HoldFirst, Unevaluated
*To*: mathgroup at smc.vnet.net
*Subject*: [mg109618] Re: HoldFirst, Unevaluated
*From*: Raffy <adraffy at gmail.com>
*Date*: Sat, 8 May 2010 07:07:38 -0400 (EDT)
*References*: <hs0pp2$dsn$1@smc.vnet.net>
On May 7, 3:24 am, Fred Klingener <gigabitbuc... at BrockEng.com> wrote:
> 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
>
> while
>
> 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?
>
> TIA,
> Fred Klingener
Function[{a, b, c}, a = Pi; 3 b + 2 c, HoldFirst]
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