39 KiB
Type Inference
Pöial's Rules
"Typing Tools for Typeless Stack Languages" by Jaanus Pöial
@INPROCEEDINGS{Pöial06typingtools,
author = {Jaanus Pöial},
title = {Typing tools for typeless stack languages},
booktitle = {In 23rd Euro-Forth Conference},
year = {2006},
pages = {40--46}
}
First Rule
This rule deals with functions (and literals) that put items on the stack (-- d):
(a -- b)∘(-- d)
---------------------
(a -- b d)
Second Rule
This rule deals with functions that consume items from the stack (a --):
(a --)∘(c -- d)
---------------------
(c a -- d)
Third Rule
The third rule is actually two rules. These two rules deal with composing functions when the second one will consume one of items the first one produces. The two types must be unified or a type conflict declared.
(a -- b t[i])∘(c u[j] -- d) t <= u (t is subtype of u)
-------------------------------
(a -- b )∘(c -- d) t[i] == t[k] == u[j]
^
(a -- b t[i])∘(c u[j] -- d) u <= t (u is subtype of t)
-------------------------------
(a -- b )∘(c -- d) t[i] == u[k] == u[j]
Examples
Let's work through some examples by hand to develop an intuition for the algorithm.
There's a function in one of the other notebooks.
F == pop swap roll< rest rest cons cons
It's all "stack chatter" and list manipulation so we should be able to deduce its type.
Stack Effect Comments
Joy function types will be represented by Forth-style stack effect comments. I'm going to use numbers instead of names to keep track of the stack arguments. (A little bit like De Bruijn index, at least it reminds me of them):
pop (1 --)
swap (1 2 -- 2 1)
roll< (1 2 3 -- 2 3 1)
These commands alter the stack but don't "look at" the values so these numbers represent an "Any type".
pop swap
(1 --) (1 2 -- 2 1)
Here we encounter a complication. The argument numbers need to be made unique among both sides. For this let's change pop to use 0:
(0 --) (1 2 -- 2 1)
Following the second rule:
(1 2 0 -- 2 1)
pop∘swap roll<
(1 2 0 -- 2 1) (1 2 3 -- 2 3 1)
Let's re-label them:
(1a 2a 0a -- 2a 1a) (1b 2b 3b -- 2b 3b 1b)
Now we follow the rules.
We must unify 1a and 3b, and 2a and 2b, replacing the terms in the forms:
(1a 2a 0a -- 2a 1a) (1b 2b 3b -- 2b 3b 1b)
w/ {1a: 3b}
(3b 2a 0a -- 2a ) (1b 2b -- 2b 3b 1b)
w/ {2a: 2b}
(3b 2b 0a -- ) (1b -- 2b 3b 1b)
Here we must apply the second rule:
(3b 2b 0a --) (1b -- 2b 3b 1b)
-----------------------------------
(1b 3b 2b 0a -- 2b 3b 1b)
Now we de-label the type, uh, labels:
(1b 3b 2b 0a -- 2b 3b 1b)
w/ {
1b: 1,
3b: 2,
2b: 3,
0a: 0,
}
(1 2 3 0 -- 3 2 1)
And now we have the stack effect comment for pop∘swap∘roll<.
Compiling pop∘swap∘roll<
The simplest way to "compile" this function would be something like:
def poswrd(s, e, d):
return roll_down(*swap(*pop(s, e, d)))
However, internally this function would still be allocating tuples (stack cells) and doing other unnecesssary work.
Looking ahead for a moment, from the stack effect comment:
(1 2 3 0 -- 3 2 1)
We should be able to directly write out a Python function like:
def poswrd(stack):
(_, (a, (b, (c, stack)))) = stack
return (c, (b, (a, stack)))
This eliminates the internal work of the first version. Because this function only rearranges the stack and doesn't do any actual processing on the stack items themselves all the information needed to implement it is in the stack effect comment.
Functions on Lists
These are slightly tricky.
rest ( [1 ...] -- [...] )
cons ( 1 [...] -- [1 ...] )
pop∘swap∘roll< rest
(1 2 3 0 -- 3 2 1) ([1 ...] -- [...])
Re-label (instead of adding left and right tags I'm just taking the next available index number for the right-side stack effect comment):
(1 2 3 0 -- 3 2 1) ([4 ...] -- [...])
Unify and update:
(1 2 3 0 -- 3 2 1) ([4 ...] -- [...])
w/ {1: [4 ...]}
([4 ...] 2 3 0 -- 3 2 ) ( -- [...])
Apply the first rule:
([4 ...] 2 3 0 -- 3 2) (-- [...])
---------------------------------------
([4 ...] 2 3 0 -- 3 2 [...])
And there we are.
pop∘swap∘roll<∘rest rest
Let's do it again.
([4 ...] 2 3 0 -- 3 2 [...]) ([1 ...] -- [...])
Re-label (the tails of the lists on each side each get their own label):
([4 .0.] 2 3 0 -- 3 2 [.0.]) ([5 .1.] -- [.1.])
Unify and update (note the opening square brackets have been omited in the substitution dict, this is deliberate and I'll explain below):
([4 .0.] 2 3 0 -- 3 2 [.0.] ) ([5 .1.] -- [.1.])
w/ { .0.] : 5 .1.] }
([4 5 .1.] 2 3 0 -- 3 2 [5 .1.]) ([5 .1.] -- [.1.])
How do we find .0.] in [4 .0.] and replace it with 5 .1.] getting the result [4 5 .1.]? This might seem hard, but because the underlying structure of the Joy list is a cons-list in Python it's actually pretty easy. I'll explain below.
Next we unify and find our two terms are the same already: [5 .1.]:
([4 5 .1.] 2 3 0 -- 3 2 [5 .1.]) ([5 .1.] -- [.1.])
Giving us:
([4 5 .1.] 2 3 0 -- 3 2) (-- [.1.])
From here we apply the first rule and get:
([4 5 .1.] 2 3 0 -- 3 2 [.1.])
Cleaning up the labels:
([4 5 ...] 2 3 1 -- 3 2 [...])
This is the stack effect of pop∘swap∘roll<∘rest∘rest.
pop∘swap∘roll<∘rest∘rest cons
([4 5 ...] 2 3 1 -- 3 2 [...]) (1 [...] -- [1 ...])
Re-label:
([4 5 .1.] 2 3 1 -- 3 2 [.1.]) (6 [.2.] -- [6 .2.])
Unify:
([4 5 .1.] 2 3 1 -- 3 2 [.1.]) (6 [.2.] -- [6 .2.])
w/ { .1.] : .2.] }
([4 5 .2.] 2 3 1 -- 3 2 ) (6 -- [6 .2.])
w/ {2: 6}
([4 5 .2.] 6 3 1 -- 3 ) ( -- [6 .2.])
First rule:
([4 5 .2.] 6 3 1 -- 3 [6 .2.])
Re-label:
([4 5 ...] 2 3 1 -- 3 [2 ...])
Done.
pop∘swap∘roll<∘rest∘rest∘cons cons
One more time.
([4 5 ...] 2 3 1 -- 3 [2 ...]) (1 [...] -- [1 ...])
Re-label:
([4 5 .1.] 2 3 1 -- 3 [2 .1.]) (6 [.2.] -- [6 .2.])
Unify:
([4 5 .1.] 2 3 1 -- 3 [2 .1.]) (6 [.2.] -- [6 .2.] )
w/ { .2.] : 2 .1.] }
([4 5 .1.] 2 3 1 -- 3 ) (6 -- [6 2 .1.])
w/ {3: 6}
([4 5 .1.] 2 6 1 -- ) ( -- [6 2 .1.])
First or second rule:
([4 5 .1.] 2 6 1 -- [6 2 .1.])
Clean up the labels:
([4 5 ...] 2 3 1 -- [3 2 ...])
And there you have it, the stack effect for pop∘swap∘roll<∘rest∘rest∘cons∘cons.
([4 5 ...] 2 3 1 -- [3 2 ...])
From this stack effect comment it should be possible to construct the following Python code:
def F(stack):
(_, (d, (c, ((a, (b, S0)), stack)))) = stack
return (d, (c, S0)), stack
Implementation
Representing Stack Effect Comments in Python
I'm going to use pairs of tuples of type descriptors, which will be integers or tuples of type descriptors:
roll_dn = (1, 2, 3), (2, 3, 1)
pop = (1,), ()
swap = (1, 2), (2, 1)
compose()
def compose(f, g):
(f_in, f_out), (g_in, g_out) = f, g
# First rule.
#
# (a -- b) (-- d)
# ---------------------
# (a -- b d)
if not g_in:
fg_in, fg_out = f_in, f_out + g_out
# Second rule.
#
# (a --) (c -- d)
# ---------------------
# (c a -- d)
elif not f_out:
fg_in, fg_out = g_in + f_in, g_out
else: # Unify, update, recur.
fo, gi = f_out[-1], g_in[-1]
s = unify(gi, fo)
if s == False: # s can also be the empty dict, which is ok.
raise TypeError('Cannot unify %r and %r.' % (fo, gi))
f_g = (f_in, f_out[:-1]), (g_in[:-1], g_out)
if s: f_g = update(s, f_g)
fg_in, fg_out = compose(*f_g)
return fg_in, fg_out
unify()
def unify(u, v, s=None):
if s is None:
s = {}
if u == v:
return s
if isinstance(u, int):
s[u] = v
return s
if isinstance(v, int):
s[v] = u
return s
return False
update()
def update(s, term):
if not isinstance(term, tuple):
return s.get(term, term)
return tuple(update(s, inner) for inner in term)
relabel()
def relabel(left, right):
return left, _1000(right)
def _1000(right):
if not isinstance(right, tuple):
return 1000 + right
return tuple(_1000(n) for n in right)
relabel(pop, swap)
(((1,), ()), ((1001, 1002), (1002, 1001)))
delabel()
def delabel(f):
s = {u: i for i, u in enumerate(sorted(_unique(f)))}
return update(s, f)
def _unique(f, seen=None):
if seen is None:
seen = set()
if not isinstance(f, tuple):
seen.add(f)
else:
for inner in f:
_unique(inner, seen)
return seen
delabel(relabel(pop, swap))
(((0,), ()), ((1, 2), (2, 1)))
C()
At last we put it all together in a function C() that accepts two stack effect comments and returns their composition (or raises and exception if they can't be composed due to type conflicts.)
def C(f, g):
f, g = relabel(f, g)
fg = compose(f, g)
return delabel(fg)
Let's try it out.
C(pop, swap)
((1, 2, 0), (2, 1))
C(C(pop, swap), roll_dn)
((3, 1, 2, 0), (2, 1, 3))
C(swap, roll_dn)
((2, 0, 1), (1, 0, 2))
C(pop, C(swap, roll_dn))
((3, 1, 2, 0), (2, 1, 3))
poswrd = reduce(C, (pop, swap, roll_dn))
poswrd
((3, 1, 2, 0), (2, 1, 3))
List Functions
Here's that trick to represent functions like rest and cons that manipulate lists. We use a cons-list of tuples and give the tails their own numbers. Then everything above already works.
rest = ((1, 2),), (2,)
cons = (1, 2), ((1, 2),)
C(poswrd, rest)
(((3, 4), 1, 2, 0), (2, 1, 4))
Compare this to the stack effect comment we wrote above:
(( (3, 4), 1, 2, 0 ), ( 2, 1, 4 ))
( [4 ...] 2 3 0 -- 3 2 [...])
The translation table, if you will, would be:
{
3: 4,
4: ...],
1: 2,
2: 3,
0: 0,
}
F = reduce(C, (pop, swap, roll_dn, rest, rest, cons, cons))
F
(((3, (4, 5)), 1, 2, 0), ((2, (1, 5)),))
Compare with the stack effect comment and you can see it works fine:
([4 5 ...] 2 3 1 -- [3 2 ...])
Dealing with cons and uncons
However, if we try to compose e.g. cons and uncons it won't work:
uncons = ((1, 2),), (1, 2)
try:
C(cons, uncons)
except Exception, e:
print e
Cannot unify (1, 2) and (1001, 1002).
unify() version 2
The problem is that the unify() function as written doesn't handle the case when both terms are tuples. We just have to add a clause to deal with this recursively:
def unify(u, v, s=None):
if s is None:
s = {}
else:
u = update(s, u)
v = update(s, v)
if u == v:
return s
if isinstance(u, int):
s[u] = v
return s
if isinstance(v, int):
s[v] = u
return s
if isinstance(u, tuple) and isinstance(v, tuple):
if len(u) != len(v) != 2:
raise ValueError(repr((u, v)))
for uu, vv in zip(u, v):
s = unify(uu, vv, s)
if s == False: # (instead of a substitution dict.)
break
return s
return False
C(cons, uncons)
((0, 1), (0, 1))
Compiling
Now consider the Python function we would like to derive:
def F_python(stack):
(_, (d, (c, ((a, (b, S0)), stack)))) = stack
return (d, (c, S0)), stack
And compare it to the input stack effect comment tuple we just computed:
F[0]
((3, (4, 5)), 1, 2, 0)
The stack-de-structuring tuple has nearly the same form as our input stack effect comment tuple, just in the reverse order:
(_, (d, (c, ((a, (b, S0)), stack))))
Remove the punctuation:
_ d c (a, (b, S0))
Reverse the order and compare:
(a, (b, S0)) c d _
((3, (4, 5 )), 1, 2, 0)
Eh?
And the return tuple
F[1]
((2, (1, 5)),)
is similar to the output stack effect comment tuple:
((d, (c, S0)), stack)
((2, (1, 5 )), )
This should make it pretty easy to write a Python function that accepts the stack effect comment tuples and returns a new Python function (either as a string of code or a function object ready to use) that performs the semantics of that Joy function (described by the stack effect.)
Python Identifiers
We want to substitute Python identifiers for the integers. I'm going to repurpose joy.parser.Symbol class for this:
from collections import defaultdict
from joy.parser import Symbol
def _names_for():
I = iter(xrange(1000))
return lambda: Symbol('a%i' % next(I))
def identifiers(term, s=None):
if s is None:
s = defaultdict(_names_for())
if isinstance(term, int):
return s[term]
return tuple(identifiers(inner, s) for inner in term)
doc_from_stack_effect()
As a convenience I've implemented a function to convert the Python stack effect comment tuples to reasonable text format. There are some details in how this code works that related to stuff later in the notebook, so you should skip it for now and read it later if you're interested.
def doc_from_stack_effect(inputs, outputs):
return '(%s--%s)' % (
' '.join(map(_to_str, inputs + ('',))),
' '.join(map(_to_str, ('',) + outputs))
)
def _to_str(term):
if not isinstance(term, tuple):
try:
t = term.prefix == 's'
except AttributeError:
return str(term)
return '[.%i.]' % term.number if t else str(term)
a = []
while term and isinstance(term, tuple):
item, term = term
a.append(_to_str(item))
try:
n = term.number
except AttributeError:
n = term
else:
if term.prefix != 's':
raise ValueError('Stack label: %s' % (term,))
a.append('.%s.' % (n,))
return '[%s]' % ' '.join(a)
compile_()
Now we can write a compiler function to emit Python source code. (The underscore suffix distiguishes it from the built-in compile() function.)
def compile_(name, f, doc=None):
if doc is None:
doc = doc_from_stack_effect(*f)
inputs, outputs = identifiers(f)
i = o = Symbol('stack')
for term in inputs:
i = term, i
for term in outputs:
o = term, o
return '''def %s(stack):
"""%s"""
%s = stack
return %s''' % (name, doc, i, o)
Here it is in action:
source = compile_('F', F)
print source
def F(stack):
"""([3 4 .5.] 1 2 0 -- [2 1 .5.])"""
(a5, (a4, (a3, ((a0, (a1, a2)), stack)))) = stack
return ((a4, (a3, a2)), stack)
Compare:
def F_python(stack):
(_, (d, (c, ((a, (b, S0)), stack)))) = stack
return ((d, (c, S0)), stack)
Next steps:
L = {}
eval(compile(source, '__main__', 'single'), {}, L)
L['F']
<function F>
Let's try it out:
from notebook_preamble import D, J, V
from joy.library import SimpleFunctionWrapper
D['F'] = SimpleFunctionWrapper(L['F'])
J('[4 5 ...] 2 3 1 F')
[3 2 ...]
With this, we have a partial Joy compiler that works on the subset of Joy functions that manipulate stacks (both what I call "stack chatter" and the ones that manipulate stacks on the stack.)
I'm probably going to modify the definition wrapper code to detect definitions that can be compiled by this partial compiler and do it automatically. It might be a reasonable idea to detect sequences of compilable functions in definitions that have uncompilable functions in them and just compile those. However, if your library is well-factored this might be less helpful.
Compiling Library Functions
We can use compile_() to generate many primitives in the library from their stack effect comments:
def defs():
roll_down = (1, 2, 3), (2, 3, 1)
roll_up = (1, 2, 3), (3, 1, 2)
pop = (1,), ()
swap = (1, 2), (2, 1)
rest = ((1, 2),), (2,)
rrest = C(rest, rest)
cons = (1, 2), ((1, 2),)
uncons = ((1, 2),), (1, 2)
swons = C(swap, cons)
return locals()
for name, stack_effect_comment in sorted(defs().items()):
print
print compile_(name, stack_effect_comment)
print
def cons(stack):
"""(1 2 -- [1 .2.])"""
(a1, (a0, stack)) = stack
return ((a0, a1), stack)
def pop(stack):
"""(1 --)"""
(a0, stack) = stack
return stack
def rest(stack):
"""([1 .2.] -- 2)"""
((a0, a1), stack) = stack
return (a1, stack)
def roll_down(stack):
"""(1 2 3 -- 2 3 1)"""
(a2, (a1, (a0, stack))) = stack
return (a0, (a2, (a1, stack)))
def roll_up(stack):
"""(1 2 3 -- 3 1 2)"""
(a2, (a1, (a0, stack))) = stack
return (a1, (a0, (a2, stack)))
def rrest(stack):
"""([0 1 .2.] -- 2)"""
((a0, (a1, a2)), stack) = stack
return (a2, stack)
def swap(stack):
"""(1 2 -- 2 1)"""
(a1, (a0, stack)) = stack
return (a0, (a1, stack))
def swons(stack):
"""(0 1 -- [1 .0.])"""
(a1, (a0, stack)) = stack
return ((a1, a0), stack)
def uncons(stack):
"""([1 .2.] -- 1 2)"""
((a0, a1), stack) = stack
return (a1, (a0, stack))
Types and Subtypes of Arguments
So far we have dealt with types of functions, those dealing with simple stack manipulation. Let's extend our machinery to deal with types of arguments.
"Number" Type
Consider the definition of sqr:
sqr == dup mul
The dup function accepts one anything and returns two of that:
dup (1 -- 1 1)
And mul accepts two "numbers" (we're ignoring ints vs. floats vs. complex, etc., for now) and returns just one:
mul (n n -- n)
So we're composing:
(1 -- 1 1)∘(n n -- n)
The rules say we unify 1 with n:
(1 -- 1 1)∘(n n -- n)
--------------------------- w/ {1: n}
(1 -- 1 )∘(n -- n)
This involves detecting that "Any type" arguments can accept "numbers". If we were composing these functions the other way round this is still the case:
(n n -- n)∘(1 -- 1 1)
--------------------------- w/ {1: n}
(n n -- )∘( -- n n)
The important thing here is that the mapping is going the same way in both cases, from the "any" integer to the number
Distinguishing Numbers
We should also mind that the number that mul produces is not (necessarily) the same as either of its inputs, which are not (necessarily) the same as each other:
mul (n2 n1 -- n3)
(1 -- 1 1)∘(n2 n1 -- n3)
-------------------------------- w/ {1: n2}
(n2 -- n2 )∘(n2 -- n3)
(n2 n1 -- n3)∘(1 -- 1 1 )
-------------------------------- w/ {1: n3}
(n2 n1 -- )∘( -- n3 n3)
Distinguishing Types
So we need separate domains of "any" numbers and "number" numbers, and we need to be able to ask the order of these domains. Now the notes on the right side of rule three make more sense, eh?
(a -- b t[i])∘(c u[j] -- d) t <= u (t is subtype of u)
-------------------------------
(a -- b )∘(c -- d) t[i] == t[k] == u[j]
^
(a -- b t[i])∘(c u[j] -- d) u <= t (u is subtype of t)
-------------------------------
(a -- b )∘(c -- d) t[i] == u[k] == u[j]
The indices i, k, and j are the number part of our labels and t and u are the domains.
By creative use of Python's "double underscore" methods we can define a Python class hierarchy of Joy types and use the issubclass() method to establish domain ordering, as well as other handy behaviour that will make it fairly easy to reuse most of the code above.
class AnyJoyType(object):
prefix = 'a'
def __init__(self, number):
self.number = number
def __repr__(self):
return self.prefix + str(self.number)
def __eq__(self, other):
return (
isinstance(other, self.__class__)
and other.prefix == self.prefix
and other.number == self.number
)
def __ge__(self, other):
return issubclass(other.__class__, self.__class__)
def __add__(self, other):
return self.__class__(self.number + other)
__radd__ = __add__
def __hash__(self):
return hash(repr(self))
class NumberJoyType(AnyJoyType): prefix = 'n'
class FloatJoyType(NumberJoyType): prefix = 'f'
class IntJoyType(FloatJoyType): prefix = 'i'
class StackJoyType(AnyJoyType):
prefix = 's'
_R = range(10)
A = map(AnyJoyType, _R)
N = map(NumberJoyType, _R)
S = map(StackJoyType, _R)
Mess with it a little:
from itertools import permutations
"Any" types can be specialized to numbers and stacks, but not vice versa:
for a, b in permutations((A[0], N[0], S[0]), 2):
print a, '>=', b, '->', a >= b
a0 >= n0 -> True
a0 >= s0 -> True
n0 >= a0 -> False
n0 >= s0 -> False
s0 >= a0 -> False
s0 >= n0 -> False
Our crude Numerical Tower of numbers > floats > integers works as well (but we're not going to use it yet):
for a, b in permutations((A[0], N[0], FloatJoyType(0), IntJoyType(0)), 2):
print a, '>=', b, '->', a >= b
a0 >= n0 -> True
a0 >= f0 -> True
a0 >= i0 -> True
n0 >= a0 -> False
n0 >= f0 -> True
n0 >= i0 -> True
f0 >= a0 -> False
f0 >= n0 -> False
f0 >= i0 -> True
i0 >= a0 -> False
i0 >= n0 -> False
i0 >= f0 -> False
Typing sqr
dup = (A[1],), (A[1], A[1])
mul = (N[1], N[2]), (N[3],)
dup
((a1,), (a1, a1))
mul
((n1, n2), (n3,))
Modifying the Inferencer
Re-labeling still works fine:
foo = relabel(dup, mul)
foo
(((a1,), (a1, a1)), ((n1001, n1002), (n1003,)))
delabel() version 2
The delabel() function needs an overhaul. It now has to keep track of how many labels of each domain it has "seen".
from collections import Counter
def delabel(f, seen=None, c=None):
if seen is None:
assert c is None
seen, c = {}, Counter()
try:
return seen[f]
except KeyError:
pass
if not isinstance(f, tuple):
seen[f] = f.__class__(c[f.prefix])
c[f.prefix] += 1
return seen[f]
return tuple(delabel(inner, seen, c) for inner in f)
delabel(foo)
(((a0,), (a0, a0)), ((n0, n1), (n2,)))
unify() version 3
def unify(u, v, s=None):
if s is None:
s = {}
else:
u = update(s, u)
v = update(s, v)
if u == v:
return s
if isinstance(u, AnyJoyType) and isinstance(v, AnyJoyType):
if u >= v:
s[u] = v
return s
if v >= u:
s[v] = u
return s
raise ValueError('Cannot unify %r and %r.' % (u, v))
if isinstance(u, tuple) and isinstance(v, tuple):
if len(u) != len(v) != 2:
raise ValueError(repr((u, v)))
for uu, vv in zip(u, v):
s = unify(uu, vv, s)
if s == False: # (instead of a substitution dict.)
break
return s
if isinstance(v, tuple):
if not stacky(u):
raise ValueError('Cannot unify %r and %r.' % (u, v))
s[u] = v
return s
if isinstance(u, tuple):
if not stacky(v):
raise ValueError('Cannot unify %r and %r.' % (v, u))
s[v] = u
return s
return False
def stacky(thing):
return thing.__class__ in {AnyJoyType, StackJoyType}
Rewrite the stack effect comments:
def defs():
roll_down = (A[1], A[2], A[3]), (A[2], A[3], A[1])
roll_up = (A[1], A[2], A[3]), (A[3], A[1], A[2])
pop = (A[1],), ()
popop = (A[2], A[1],), ()
popd = (A[2], A[1],), (A[1],)
popdd = (A[3], A[2], A[1],), (A[2], A[1],)
swap = (A[1], A[2]), (A[2], A[1])
rest = ((A[1], S[1]),), (S[1],)
rrest = C(rest, rest)
cons = (A[1], S[1]), ((A[1], S[1]),)
ccons = C(cons, cons)
uncons = ((A[1], S[1]),), (A[1], S[1])
swons = C(swap, cons)
dup = (A[1],), (A[1], A[1])
dupd = (A[2], A[1]), (A[2], A[2], A[1])
mul = (N[1], N[2]), (N[3],)
sqrt = C(dup, mul)
first = ((A[1], S[1]),), (A[1],)
second = C(rest, first)
third = C(rest, second)
tuck = (A[2], A[1]), (A[1], A[2], A[1])
over = (A[2], A[1]), (A[2], A[1], A[2])
succ = pred = (N[1],), (N[2],)
divmod_ = pm = (N[2], N[1]), (N[4], N[3])
return locals()
DEFS = defs()
for name, stack_effect_comment in sorted(DEFS.items()):
print name, '=', doc_from_stack_effect(*stack_effect_comment)
ccons = (a0 a1 [.0.] -- [a0 a1 .0.])
cons = (a1 [.1.] -- [a1 .1.])
divmod_ = (n2 n1 -- n4 n3)
dup = (a1 -- a1 a1)
dupd = (a2 a1 -- a2 a2 a1)
first = ([a1 .1.] -- a1)
mul = (n1 n2 -- n3)
over = (a2 a1 -- a2 a1 a2)
pm = (n2 n1 -- n4 n3)
pop = (a1 --)
popd = (a2 a1 -- a1)
popdd = (a3 a2 a1 -- a2 a1)
popop = (a2 a1 --)
pred = (n1 -- n2)
rest = ([a1 .1.] -- [.1.])
roll_down = (a1 a2 a3 -- a2 a3 a1)
roll_up = (a1 a2 a3 -- a3 a1 a2)
rrest = ([a0 a1 .0.] -- [.0.])
second = ([a0 a1 .0.] -- a1)
sqrt = (n0 -- n1)
succ = (n1 -- n2)
swap = (a1 a2 -- a2 a1)
swons = ([.0.] a0 -- [a0 .0.])
third = ([a0 a1 a2 .0.] -- a2)
tuck = (a2 a1 -- a1 a2 a1)
uncons = ([a1 .1.] -- a1 [.1.])
globals().update(DEFS)
Compose dup and mul
C(dup, mul)
((n0,), (n1,))
Revisit the F function, works fine.
F = reduce(C, (pop, swap, roll_down, rest, rest, cons, cons))
F
(((a0, (a1, s0)), a2, a3, a4), ((a3, (a2, s0)),))
print doc_from_stack_effect(*F)
([a0 a1 .0.] a2 a3 a4 -- [a3 a2 .0.])
Some otherwise inefficient functions are no longer to be feared. We can also get the effect of combinators in some limited cases.
def neato(*funcs):
print doc_from_stack_effect(*reduce(C, funcs))
# e.g. [swap] dip
neato(roll_up, swap, roll_down)
(a0 a1 a2 -- a1 a0 a2)
# e.g. [popop] dip
neato(popdd, roll_down, pop)
(a0 a1 a2 a3 -- a2 a3)
# Reverse the order of the top three items.
neato(roll_up, swap)
(a0 a1 a2 -- a2 a1 a0)
compile_() version 2
Because the type labels represent themselves as valid Python identifiers the compile_() function doesn't need to generate them anymore:
def compile_(name, f, doc=None):
inputs, outputs = f
if doc is None:
doc = doc_from_stack_effect(inputs, outputs)
i = o = Symbol('stack')
for term in inputs:
i = term, i
for term in outputs:
o = term, o
return '''def %s(stack):
"""%s"""
%s = stack
return %s''' % (name, doc, i, o)
print compile_('F', F)
def F(stack):
"""([a0 a1 .0.] a2 a3 a4 -- [a3 a2 .0.])"""
(a4, (a3, (a2, ((a0, (a1, s0)), stack)))) = stack
return ((a3, (a2, s0)), stack)
But it cannot magically create new functions that involve e.g. math and such. Note that this is not a sqr function implementation:
print compile_('sqr', C(dup, mul))
def sqr(stack):
"""(n0 -- n1)"""
(n0, stack) = stack
return (n1, stack)
compilable()
The functions that can be compiled are the ones that have only AnyJoyType and StackJoyType labels in their stack effect comments. We can write a function to check that:
from itertools import imap
def compilable(f):
return isinstance(f, tuple) and all(imap(compilable, f)) or stacky(f)
for name, stack_effect_comment in sorted(defs().items()):
if compilable(stack_effect_comment):
print name, '=', doc_from_stack_effect(*stack_effect_comment)
ccons = (a0 a1 [.0.] -- [a0 a1 .0.])
cons = (a1 [.1.] -- [a1 .1.])
dup = (a1 -- a1 a1)
dupd = (a2 a1 -- a2 a2 a1)
first = ([a1 .1.] -- a1)
over = (a2 a1 -- a2 a1 a2)
pop = (a1 --)
popd = (a2 a1 -- a1)
popdd = (a3 a2 a1 -- a2 a1)
popop = (a2 a1 --)
rest = ([a1 .1.] -- [.1.])
roll_down = (a1 a2 a3 -- a2 a3 a1)
roll_up = (a1 a2 a3 -- a3 a1 a2)
rrest = ([a0 a1 .0.] -- [.0.])
second = ([a0 a1 .0.] -- a1)
swap = (a1 a2 -- a2 a1)
swons = ([.0.] a0 -- [a0 .0.])
third = ([a0 a1 a2 .0.] -- a2)
tuck = (a2 a1 -- a1 a2 a1)
uncons = ([a1 .1.] -- a1 [.1.])
Functions that use the Stack
Consider the stack function which grabs the whole stack, quotes it, and puts it on itself:
stack (... -- ... [...] )
stack (... a -- ... a [a ...] )
stack (... b a -- ... b a [a b ...])
We would like to represent this in Python somehow. To do this we use a simple, elegant trick.
stack S -- ( S, S)
stack (a, S) -- ( (a, S), (a, S))
stack (a, (b, S)) -- ( (a, (b, S)), (a, (b, S)))
Instead of representing the stack effect comments as a single tuple (with N items in it) we use the same cons-list structure to hold the sequence and unify() the whole comments.
stack∘uncons
Let's try composing stack and uncons. We want this result:
stack∘uncons (... a -- ... a a [...])
The stack effects are:
stack = S -- (S, S)
uncons = ((a, Z), S) -- (Z, (a, S))
Unifying:
S -- (S, S) ∘ ((a, Z), S) -- (Z, (a, S ))
w/ { S: (a, Z) }
(a, Z) -- ∘ -- (Z, (a, (a, Z)))
So:
stack∘uncons == (a, Z) -- (Z, (a, (a, Z)))
It works.
stack∘uncons∘uncons
Let's try stack∘uncons∘uncons:
(a, S ) -- (S, (a, (a, S ))) ∘ ((b, Z), S` ) -- (Z, (b, S` ))
w/ { S: (b, Z) }
(a, (b, Z)) -- ((b, Z), (a, (a, (b, Z)))) ∘ ((b, Z), S` ) -- (Z, (b, S` ))
w/ { S`: (a, (a, (b, Z))) }
(a, (b, Z)) -- ((b, Z), (a, (a, (b, Z)))) ∘ ((b, Z), (a, (a, (b, Z)))) -- (Z, (b, (a, (a, (b, Z)))))
(a, (b, Z)) -- (Z, (b, (a, (a, (b, Z)))))
It works.
compose() version 2
This function has to be modified to use the new datastructures and it is no longer recursive, instead recursion happens as part of unification.
def compose(f, g):
(f_in, f_out), (g_in, g_out) = f, g
if not g_in:
fg_in, fg_out = f_in, stack_concat(g_out, f_out)
elif not f_out:
fg_in, fg_out = stack_concat(f_in, g_in), g_out
else: # Unify and update.
s = unify(g_in, f_out)
if s == False: # s can also be the empty dict, which is ok.
raise TypeError('Cannot unify %r and %r.' % (fo, gi))
fg_in, fg_out = update(s, (f_in, g_out))
return fg_in, fg_out
stack_concat = lambda q, e: (q[0], stack_concat(q[1], e)) if q else e
I don't want to rewrite all the defs myself, so I'll write a little conversion function instead. This is programmer's laziness.
def sequence_to_stack(seq, stack=StackJoyType(23)):
for item in seq: stack = item, stack
return stack
NEW_DEFS = {
name: (sequence_to_stack(i), sequence_to_stack(o))
for name, (i, o) in DEFS.iteritems()
}
globals().update(NEW_DEFS)
stack = S[0], (S[0], S[0])
C(stack, uncons)
((a0, s0), (s0, (a0, (a0, s0))))
C(C(stack, uncons), uncons)
((a0, (a1, s0)), (s0, (a1, (a0, (a0, (a1, s0))))))
The display function should be changed too.
doc_from_stack_effect() version 2
Clunky junk, but it will suffice for now.
def doc_from_stack_effect(inputs, outputs):
switch = [False] # Do we need to display the '...' for the rest of the main stack?
i, o = _f(inputs, switch), _f(outputs, switch)
if switch[0]:
i.append('...')
o.append('...')
return '(%s--%s)' % (
' '.join(reversed([''] + i)),
' '.join(reversed(o + [''])),
)
def _f(term, switch):
a = []
while term and isinstance(term, tuple):
item, term = term
a.append(item)
assert isinstance(term, StackJoyType), repr(term)
a = [_to_str(i, term, switch) for i in a]
return a
def _to_str(term, stack, switch):
if not isinstance(term, tuple):
if term == stack:
switch[0] = True
return '[...]'
return (
'[.%i.]' % term.number
if isinstance(term, StackJoyType)
else str(term)
)
a = []
while term and isinstance(term, tuple):
item, term = term
a.append(_to_str(item, stack, switch))
assert isinstance(term, StackJoyType), repr(term)
if term == stack:
switch[0] = True
end = '...'
else:
end = '.%i.' % term.number
a.append(end)
return '[%s]' % ' '.join(a)
for name, stack_effect_comment in sorted(NEW_DEFS.items()):
print name, '=', doc_from_stack_effect(*stack_effect_comment)
ccons = (a0 a1 [.0.] -- [a0 a1 .0.])
cons = (a1 [.1.] -- [a1 .1.])
divmod_ = (n2 n1 -- n4 n3)
dup = (a1 -- a1 a1)
dupd = (a2 a1 -- a2 a2 a1)
first = ([a1 .1.] -- a1)
mul = (n1 n2 -- n3)
over = (a2 a1 -- a2 a1 a2)
pm = (n2 n1 -- n4 n3)
pop = (a1 --)
popd = (a2 a1 -- a1)
popdd = (a3 a2 a1 -- a2 a1)
popop = (a2 a1 --)
pred = (n1 -- n2)
rest = ([a1 .1.] -- [.1.])
roll_down = (a1 a2 a3 -- a2 a3 a1)
roll_up = (a1 a2 a3 -- a3 a1 a2)
rrest = ([a0 a1 .0.] -- [.0.])
second = ([a0 a1 .0.] -- a1)
sqrt = (n0 -- n1)
succ = (n1 -- n2)
swap = (a1 a2 -- a2 a1)
swons = ([.0.] a0 -- [a0 .0.])
third = ([a0 a1 a2 .0.] -- a2)
tuck = (a2 a1 -- a1 a2 a1)
uncons = ([a1 .1.] -- a1 [.1.])
print ; print doc_from_stack_effect(*stack)
print ; print doc_from_stack_effect(*C(stack, uncons))
print ; print doc_from_stack_effect(*C(C(stack, uncons), uncons))
print ; print doc_from_stack_effect(*C(C(stack, uncons), cons))
(... -- ... [...])
(... a0 -- ... a0 a0 [...])
(... a1 a0 -- ... a1 a0 a0 a1 [...])
(... a0 -- ... a0 [a0 ...])
print doc_from_stack_effect(*C(ccons, stack))
(... a1 a0 [.0.] -- ... [a1 a0 .0.] [[a1 a0 .0.] ...])
Q = C(ccons, stack)
Q
((s0, (a0, (a1, s1))), (((a1, (a0, s0)), s1), ((a1, (a0, s0)), s1)))
compile_() version 3
This makes the compile_() function pretty simple as the stack effect comments are now already in the form needed for the Python code:
def compile_(name, f, doc=None):
i, o = f
if doc is None:
doc = doc_from_stack_effect(i, o)
return '''def %s(stack):
"""%s"""
%s = stack
return %s''' % (name, doc, i, o)
print compile_('Q', Q)
def Q(stack):
"""(... a1 a0 [.0.] -- ... [a1 a0 .0.] [[a1 a0 .0.] ...])"""
(s0, (a0, (a1, s1))) = stack
return (((a1, (a0, s0)), s1), ((a1, (a0, s0)), s1))
unstack = (S[1], S[0]), S[1]
enstacken = S[0], (S[0], S[1])
print doc_from_stack_effect(*unstack)
([.1.] --)
print doc_from_stack_effect(*enstacken)
(-- [.0.])
print doc_from_stack_effect(*C(cons, unstack))
(a0 [.0.] -- a0)
print doc_from_stack_effect(*C(cons, enstacken))
(a0 [.0.] -- [[a0 .0.] .1.])
C(cons, unstack)
((s0, (a0, s1)), (a0, s0))
Sets of Stack Effects
...
concat
How to deal with concat?
concat ([.0.] [.1.] -- [.0. .1.])
We would like to represent this in Python somehow...
concat = (S[0], S[1]), ((S[0], S[1]),)
But this is actually cons with the first argument restricted to be a stack:
([.0.] [.1.] -- [[.0.] .1.])
What we have implemented so far would actually only permit:
([.0.] [.1.] -- [.2.])
concat = (S[0], S[1]), (S[2],)
Which works but can lose information. Consider cons concat, this is how much information we could retain:
(1 [.0.] [.1.] -- [1 .0. .1.])
As opposed to just:
(1 [.0.] [.1.] -- [.2.])
Which works but can lose information. Consider cons concat, this is how much information we could retain:
(1 [.0.] [.1.] -- [1 .0. .1.]) uncons uncons
(1 [.0.] [.1.] -- 1 [.0. .1.]) uncons
So far so good...
(1 [2 .2.] [.1.] -- 1 2 [.2. .1.])
(1 [.0.] [.1.] -- 1 [.0. .1.]) ([a1 .10.] -- a1 [.10.])
w/ { [a1 .10.] : [ .0. .1.] }
-or-
w/ { [ .0. .1.] : [a1 .10. ] }
Typing Combinators
TBD
This is an open subject.
The obvious thing is that you now need two pairs of tuples to describe the combinators' effects, a stack effect comment and an expression effect comment:
dip (a [F] --)--(-- F a)
One thing that might help is...
Abstract Interpretation
Something else...
[4 5 ...] 2 1 0 pop∘swap∘roll<∘rest∘rest∘cons∘cons
[4 5 ...] 2 1 swap∘roll<∘rest∘rest∘cons∘cons
[4 5 ...] 1 2 roll<∘rest∘rest∘cons∘cons
1 2 [4 5 ...] rest∘rest∘cons∘cons
1 2 [5 ...] rest∘cons∘cons
1 2 [...] cons∘cons
1 [2 ...] cons
[1 2 ...]
Eh?