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RE: Help: Why no output?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg34956] RE: [mg34941] Help: Why no output?
*From*: "DrBob" <majort at cox-internet.com>
*Date*: Sat, 15 Jun 2002 02:27:41 -0400 (EDT)
*Reply-to*: <drbob at bigfoot.com>
*Sender*: owner-wri-mathgroup at wolfram.com
By the way, a solution is to use a definition like this:
ClearAll[f]
f[x_, n__Integer] := If[Length[{n}] == 1, n , Length[x]]
f[x_]:=Length[x]
(two underscores where you had three)
Even better, you could use
ClearAll[f]
f[x_, n_Integer] := n
f[x_, n___Integer] := Length[x]
(one underscore in the first rule, three in the second)
Bobby
-----Original Message-----
From: DrBob [mailto:majort at cox-internet.com]
To: mathgroup at smc.vnet.net
Subject: [mg34956] RE: [mg34941] Help: Why no output?
Kezhao,
When you give no second argument to f, n is Sequence[]... a list of
arguments to place in some function, but a list of zero length!
Hence in that case
If[Length[{n}] == 1, n, Length[x]]
is just:
If[Length[{}] == 1, Length[x]]
(NO arguments are inserted at the position of n.)
Now, also
1*Sequence[]
resolves to
Times[1, Sequence[]]
and hence to
Times[1]
and hence to
1
Therefore, when n=Sequence[], the following
If[Length[{n}] == 1, 1*n, Length[x]]
resolves to
If[Length[{}] == 1, 1, Length[x]]
which is just
Length[x]
Bobby Treat
-----Original Message-----
From: Kezhao Zhang [mailto:kzhang at flashmail.com]
To: mathgroup at smc.vnet.net
Subject: [mg34956] [mg34941] Help: Why no output?
The behavior of the following function is puzzling to me:
In[1]:=f[x_, n___Integer] := If[Length[{n}] == 1, n , Length[x]]
In[2]:=f[0.5] (* Nothing returned. *)
In[3]:=f[{1,2}] (* Nothing returned. The length of x should be
returned *)
(* However, with the modification to f, everything works as intended
*)
In[4]:=f[x_, n___Integer] := If[Length[{n}] == 1, 1*n , Length[x]]
^^^^^^
In[5]:=f[0.5]
Out[5]:=0
In[6]:=f[{1,2}]
Out[6]:=2
Could anyone help me understand why changing n to (1*n) in the If[]
statement makes such difference? Why doesn't f[] defined in the In[1]
work?
Thanks for your help.
Kezhao Zhang
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