I don't have formal knowledge of continuations, and am wondering if someone can help me verify and understand the code I wrote :).
Problem
The general problem I'm trying to solve is to convert expressions like
(2 * var) + (3 * var) == 4
into functions
\x y -> 2 * x + 3 * y == 4 -- (result)
which can be then passed into the yices-painless package.
Motivation
As a simpler example, note that var is translated into \x -> x. How can we multiply two var's (denote them \x -> x and \y -> y) into one expression \x -> \y -> x * y?
I've heard continuations described as the "rest of computation", and thought that's what I need. Following that idea, a var should take a function
f :: α -> E -- rest of computation
whose argument will be the value of the variable var created, and return what we want (code listing marked result), a new function taking a variable x and returning f x. Hence, we define,
var' = \f -> (\x -> f x)
Then, for multiplication, say of xf and yf (which could be equal to var, for example), we want to take a "rest of computation" function f :: α -> E as above, and return a new function. We know what the function should do given the values of xf and yf (denoted x and y below), and define it as so,
mult xf yf = \f -> xf (\x -> yf (\y -> f (x Prelude.* y)))
Code
const' c = \f -> f c
var' = \f -> (\x -> f x) -- add a new argument, "x", to the function
add xf yf = \f -> xf (\x -> yf (\y -> f (x Prelude.+ y)))
mult xf yf = \f -> xf (\x -> yf (\y -> f (x Prelude.* y)))
v_α = var' -- "x"
v_β = var' -- "y"
v_γ = var' -- "z"
m = mult v_α v_β -- "x * y"
a = add m v_γ -- "x * y + z"
eval_six = (m id) 2 3
eval_seven = (a id) 2 3 1
two = const' 2 -- "2"
m2 = mult two v_γ -- "2 * z"
a2 = add m m2 -- "x * y + 2 * z"
eval_two = (m2 id) 1
eval_eight = (a2 id) 2 3 1
quad_ary = (var' `mult` var') `mult` (var' `mult` var')
eval_thirty = (quad_ary id) 1 2 3 5
well, it seems to work.
