On 19/01/2013, at 2:11 PM, john skaller wrote:
>
> If I use the type int ^ (2 + 5 +2) at present the code doesn't work.
this does work if you get the @#$%^&*( syntax right:
var x = (1,2,3,4,5,6,7,8,9);
var z = x :>> (int ^ (2 + 5 + 2));
println$ z;
fun f(x:5) => (case 1 of (2 + 5 + 2)) x;
fun g(y:5):int => f y z;
var a = f (1:>>5) z;
var b = g (1:>>5);
println$ a;
println$ b;
So here the function "g" is effectively an array representing
the slice 3 .. 7.
However you cannot retrieve it as an object, I guess because
it's an indexing formula for a linear array, so there isn't
actually any slice.
The coercion here:
var x = (1,2,3,4,5,6,7,8,9);
var y = x :>> (int ^2*int^5*int^2);
println$ y;
reshapes the array, the coercion to z reshapes
the index. The reshaped array requires two indexing
operations to get an element:
y . 1 . 1
Using the reshaped index that's
((case 1 of (2 + 5 + 2)) (case 1 of 5)) z;
i.e. first select the middle 5 elements, then the second (index 1)
element of that.
Note you can do this:
println$ z . ( (case 1 of 5) . (case 1 of (2 + 5 + 2)));
but you cannot remove the parens after z . so this kind of
reverse application isn't associative. Hmm.
Note: with something like
case 1 of 5
you'd like to write:
case j of 5
where j was a variable (or expression) but you cannot. That's because
those case things are literal constants. Indeed
case 1 of (2 + 5 + 2)
appears to be applied to case 1 of 5, but this is not so: it's a constructor
not a function. Felix can wrap constructors into functions though.
So if you want to make a "variable" case of type 5 here's how:
match j with
| 0 => case 0 of 5
| 1 => case 1 of 5
...
of course here j is an int. A quick way to do the above is:
j :>> 5
which is basically
j modulo 5
in "mathematical" notation. Note that here "modulo" is NOT an operator,
just as in maths:
j % 5 is an int
j modulo 5 is a member of the cyclic group Z_5
At present in Felix, the type 5 is not Z_5, rather it's just the underlying
set: there are NO operations on values of type 5.
Of course it would be interesting to make 5 equal to Z_5, the clock
with 5 digits, then we would have addition. The problem here is that
4 + 4 = 3
that is, there's no error from overflow. This has + and - as usual.
The classic useful case is the formula for a determinant, where
it is cool if the indices are modulo N, so when you increment
them you just wrap around in the matrix. but it also means
you can index past the end of an array and wrap around
to the start, without realising your code was bugged,
in the case you really didn't want that.
Anyhow I am thinking to add the notation
expression modulo base
as a way to build modular numbers: base has to be a constant.
It means, do the calculation of expression in Z_base.
This is hard because whilst you can just do addition and multiplication
and subtraction with ints and % base after, division doesn't work that way.
--
john skaller
skal...@users.sourceforge.net
http://felix-lang.org
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