How to use local variables with indices in a Mathematica Block environment?

Question

I have the following code:

Subscript[f, i_][x_] := Block[{a},
  Subscript[a, i] = 3 x;
  Subscript[a, 1]
]

If I evaluate for example Subscript[f,1][2], it evaluates to 6. After that asking what Subscript[a,1] is also gives me six. However, I want a to be a local variable.

My question: how can I define a local variable while using indices...?

Note: the original post mentioned that evaluating Subscript[f,1][2] = 6 would lead to the described behaviour, which just isn't true. A reasonable guess is that this was ment to mean that Subscript[f,1][2] was to be evaluated and it would evaluate to 6 (which it does). After that Subscript[a,1] will evaluate to 6 as reported.

Answer 1

The problem is that the definition which makes Subscript[a,i] evaluate to 6 is stored in the DownValues of Subscript. The down values of Subscript don't care about the blocking of a and will still exist when Block exits. You can check that by evaluating DownValues[Subscript]. You could use TagSet (shortcut: /:) to store that value in the UpValues of the localized a, which would behave as you expect (i.e. the rule in UpValues of a is removed when Block exits):

ClearAll[f, a, Subscript]
Subscript[f, i_][x_] := Block[{a},
  a /: Subscript[a, i] = 3 x;
  Subscript[a, 1]
]

on the other hand I wouldn't actually recommend to use such subscripted variables, what you have seen is probably only one of several unexpected things that could happen.

Side Note:It probably should be mentioned that the definition for Subscript[f,i_][x_] is also not associated with f but with Subscript, which can be checked by evaluating SubValues[Subscript]. Unfortunately this can not be changed by using TagSet for f, since Mathematica here will complain about f being too deep in the left hand side expression. One possibility to change that is to reformulate the function definition to e.g.:

f /: Subscript[f, i_] = Function[x,
  Block[{a},
   a /: Subscript[a, i] = 3 x;
   Subscript[a, 1]
   ]
  ]

which will behave the same as the original in most practical cases but of course is not exactly the same thing. You can use UpValues[f] to check that this rule is now really associated with f.

Answer 2

I think you want to use Module[] instead of Block[]

Subscript[f, i_][x_] := Module[{a}, Subscript[a, i] = 3 x; Subscript[a, 1]]

Answer 3

As Albert Retey says, the issue is that

Subscript[a, i] = 3 x

affects the DownValues of Subscript, not any of the values of a. Subscript is one of the few system-defined symbols where you can modify its values without unprotecting it first (Derivative is another). Thus, in order to localize changes to subscripts of a, you need to use Block on Subscript instead of a. Thus, for your simple use case, you can do:

Subscript[h, i_][x_] := Block[{Subscript},
  Subscript[b, i] = 3 x;
  Subscript[b, i]]

Now things work as you want:

In[17]:= Subscript[h, 4][3]
Out[17]= 9

In[18]:= Subscript[b, 3]
Out[18]= Subscript[b, 3]

This has the drawback that you lose , because you lose all access to values of Subscript from outside the Block. For a somewhat contrived example of what can go wrong, consider:

Subscript[c, 1] = 17;

Subscript[hh, i_][x_] := Block[{Subscript},
  Subscript[b, i] = If[OddQ@Subscript[c, 1], 3 x, 4 x];
  Subscript[b, i]]

In[23]:= Subscript[hh, 4][17]
Out[23]= 68

In[24]:= OddQ@Subscript[c, 1]
Out[24]= True

This is perhaps not what you wanted to have happen. Now you can go ahead and manipulate things so that Subscript gets all the DownValues and SubValues from outside the Block, but this is an error-prone and likely inefficient hassle. However, there's an internal function that does exactly what you would want in this case, called InternalInheritedBlock`:

Subscript[hhh, i_][x_] := Internal`InheritedBlock[{Subscript},
  Subscript[b, i] = If[OddQ@Subscript[c, 1], 3 x, 4 x];
  Subscript[b, i]]

Now you get the subscripts you want from outside without having to worry about clobbering anything:

In[31]:= Subscript[hhh, 4][17]
Out[31]= 51

In[32]:= Subscript[b, 3]
Out[32]= Subscript[b, 3]