After playing with Mathematica's symbolic and numerical capabilities, I find it to be a decent programming language, too. However, something making it less appealing as a general-purpose language is the lack of C-like struct data type (or the record type as known in Pascal). How can I get around this problem?
+5
A:
If I understand your question correctly, you can simply write things like this:
x[foo] = bar x[bar] = baz x[1] = 7 x[7] = 1 ?x
Then to access the data for any specific index just type the same (e.g., x[1] will return 7, x[foo] will return bar).
Will Robertson
2009-09-21 06:42:41
This way of doing things has one real advantage over the rule approach suggested by sakra: it allows you to mutate the "fields" of the "struct" in a straightforward way.
Pillsy
2009-09-22 13:20:09
There's a problem with your answer: if a "field" is a list, its elements can't be changed individually. For example,x[foo]={1,2};x[foo] [[1]] =3 (* attempting to change list element *)will result in an error, because x[foo] is not an Lvalue.So it still doesn't fully replace C struct functionality.
2009-09-23 01:43:32
+2
A:
You can use a Mathematica rule lists to mimic a C-like struct data type. E.g.,:
person = {firstName -> "John", lastName -> "Doe"}
You can then access the record's fields by using the /. operator:
firstName /. person
yields John.
lastName /. person
yields Doe.
Also see the Mathematica documentation on transformation rules.
sakra
2009-09-22 11:17:49
+1
A:
The transformation rules approach has the same drawback as the bracket approach. For instance, suppose you tried: (lastName /. person) = "Harvey" The result would be an error, not an assignment. The method works for retrieval but not for storage.
Jeff Brown at MSU Economics
2009-10-24 04:47:50
+1
A:
This way can work:
x[foo] = bar
x[bar] = baz
x[1] = 7
x[7] = 1
x[c] = {{1,2,3},{4,5,6}}
and also for changing the elements of a list field you can so the following:
x[c] = ReplacePart[x[c], {1, 1} -> 8]
which returns:
x[c] = {{8,2,3},{4,5,6}}
Atefeh
2010-02-05 11:20:57