Modularity and the Naming of Things
- To: mathgroup at smc.vnet.net
- Subject: [mg38027] Modularity and the Naming of Things
- From: Name <mee at home.com.redline.ru>
- Date: Tue, 26 Nov 2002 00:49:18 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
1. Consider
In[1]:=
Function[y,Module[{x=1},y+x]][x]
Module[{x=1},#+x]&[x]
Out[1]=
1+x
Out[2]=
2
The first example always works as a function returning its argument plus
one (unless we pass it 'unallowed' argument x$); the second example
doesn't.
I think the unnamed Slot parameters should internally be treated this
way:
In[3]:=
Function[slot1,Module[{x=1},slot1+x]][x]
Out[3]=
1+x
This works consistently for any arguments passed to the function,
doesn't interfere with the Module local variables and gives the same
results as the
definition using named parameters.
2. Consider
In[4]:=
x=1;
Function[x,Evaluate@x]
Module[{x=2},Function[x,Evaluate@x]]
x=.
Out[5]=
Function[x,1]
Out[6]=
Function[x,1]
The reason why the second example works this way is that Function makes
the global x variable local, not the Module's x$n.
This means we can't define functions that depend on Module's local
variables unless we use some special tricks.
Wouldn't it be better if Function made Module's variables local instead
so that
Module[{x}, Function[x,fx]]
became
Module[{x}, Function[x$n,fx]]?
This behaviour would be useful in conjunction with the modification
discussed below. The same goes for Rule and Set inside Module.
3. Consider
In[8]:=
y=x;
Function[x,Evaluate[2x+y]]
y=.
Module[{y=x},
Function[x,Evaluate[2x+y]]]
Out[9]=
Function[x,3 x]
Out[11]=
Function[x$,x+2 x$]
problem is that it makes the task of creating functions inside Modules
much more complicated;
if we want to do something like Unapply[fx, x], there is a simple
solution Evaluate[fx/.x->#]& but it doesn't work for fx=#+x&.
Is it possible to change the evaluation rules so that in
Function[x,Evaluate[fx]] fx is evaluated outside the Function scope?
Then the result is passed to the Function; in the second example above
it would work as follows:
Evaluate[2x+y]->3x->Function[x,3x],
or, in the case of x being local to the Module,
Evaluate[2x$n+y]->3x$n->Function[x$n,3x$n].
The same goes for the right-hand side of Rule and Set.
Maxim Rytin
m.r at prontomail.com