Gustavo

Introduction

Looking for an example

#lang racket

(define (-check-non-negative-integer who i false-ok?)
  (when (or i (not false-ok?))
    (unless (and (integer? i) (exact? i) (not (negative? i)))
      (raise-argument-error [...]))))

(define (check-non-negative-integer who i)
  (-check-non-negative-integer who i #f))

(define (check-non-negative-integer/false who i)
  (-check-non-negative-integer who i #t))

#lang racket
  
{define (check my-false)
  (if (if (random 1) #t my-false) 2 3)}

(check #f)

#lang racket

{define check
  (#%closed check
    {lambda (my-false)
      (if (if (random 1) #t my-false) 2 3)})}

(print-values (if (random 1) (if #t 2 3) 3))

The origin of the problem

(if #t X Y)  -->  X
(if #f X Y)  -->  Y
(if (not P) X Y)  -->  (if P Y X)

Old and new optimizations

a: [boolean] (if M #t #f)  -->  M
b: [boolean] (if v v X)  -->  (if v #t X)
c: [new] (if v X v)  -->  (if v X #f)

d: (if (not P) X Y)  -->  (if P Y X)
d: (if #t X Y)  -->  X
e: (if #f X Y)  -->  Y

f: (if v v #f)  -->  v
g: (if <omitable> v v)  -->  v

h: (if (if M N #f) P K) => (if M (if N P K) K)
i: [new] (if M (if (not Q) P K) K) => (if M (if Q K P) K)
j: [new] (if M (if #t P K) K) => (if M P K)
k: [new] (if M (if #f P K) K) => (if M K K)

  /* Convert (if <id> expr <id>) to (if <id> expr #f) */
  if (SAME_TYPE(SCHEME_TYPE(t), scheme_local_type)
      && SAME_TYPE(SCHEME_TYPE(fb), scheme_local_type)
      && (SCHEME_LOCAL_POS(t) == SCHEME_LOCAL_POS(fb))) {
    b->fbranch = fb = scheme_false;
  }
  
  if (context & OPT_CONTEXT_BOOLEAN) {
    /* Convert (if <id> <id> expr) to (if <id> #t expr) */
    if (SAME_TYPE(SCHEME_TYPE(t), scheme_local_type)
        && SAME_TYPE(SCHEME_TYPE(tb), scheme_local_type)
        && (SCHEME_LOCAL_POS(t) == SCHEME_LOCAL_POS(tb))) {
      b->tbranch = tb = scheme_true;
    }

    /* For test position, convert (if <expr> #t #f) to <expr> */
    if (SAME_OBJ(tb, scheme_true) && SAME_OBJ(fb, scheme_false))
      return scheme_optimize_expr(t, info, context);
  }

Versions to compile and test

Results of a-b-c

• just "a"
• with "a-b", this is code that uses the unmodified Racket
• with "b-a", use the reverse order
• "c-b-a", use the reverse order and add "c"

• The optimization "b" transforms 170 appearances (if v v ?) into (if v # t ?).
• Similarly, the optimization "c" transforms about 260 appearances (if v ? v) into (if v ? #f).
• I expected that swapping the optimizations "a" and "b" they could compose, so that there is a transformation like (if v v #f) into (if v #t #f) into v, but there are not too many of these cases.
• (If we eliminate the optimization "a", there appear approximately 5000 "new" instances of (if ? #t #f), it's a very bad idea.)

Results of h-i-j-k

• without "h"
• with "h" , this is the code that uses unmodified Racket
• with "h-i-j-k ", i.e. optimizing the result of "h"

• With regard to (if #t ?) and (if #f ?), comparing different versions, we see that the optimization "h" is causing them. The if with #t and #f don't appear when "h" is not present and they disappear when the result of "h" is further optimized.
• However, there are several instances of (if (not ?) ? ?) in all the versions. Moreover, after optimizing the result of "h" not all the new instances disappear. It'd like investigate their origin and see how they can be eliminated, because they appear more than 1500 times.

Ideas and remarks

• It would have been better to use the optimizer internal function scheme_optimize_expr or optimize_branch to optimize the second if in the result of the optimization "h". It would be more elegant because we wouldn't need to repeat the code and this alternative method would apply all the other optimizations, although it would be slightly slower. The problem is that to call these functions we must use a Optimize_Info structure containing information about the state of the program being optimized . I tried several times, but I could not write it correctly.
• There are many (if ? #f #t), maybe they can be replaced by (not ?). I did some microbenchmarks to measure the speed difference, but I saw almost no change. However, the modification might allow the optimizer to apply other optimizations. Something similar happens with (if ? #t #f), but I don't know what would be the name of the associated function.
• There are 31 occurrences of (if (if ? #<void> ?) ?) and 11 occurrences of (if (if ? '() #f) ?). It's possible to add a optimization for them, but they appear very seldom so probably it's not very useful.
• In each recompilation, the number of times that each thing appears changes a little. It's very strange. It doesn't affect the results too much, but we must be careful when comparing very close  values.
• There is exactly one (if (if ? 1 0) ? ?). I did not seek the code that produces it, but at first glance it looks like a (if (bool->binary ?) ? ?), if bool->binary were a real function. Is this a bug or just a strange coincidence produced by all the intermediate expansions?