All ready to implement property lists, not yet done.

This commit is contained in:
Simon Brooke 2023-04-02 03:45:40 +01:00
parent 5ee9531e6b
commit b61e7c3e8c
16 changed files with 225 additions and 125 deletions

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@ -1,21 +0,0 @@
;; see page 70 of Lisp 1.5 Programmers Manual; this expands somewhat
;; on the accounts of eval and apply given on page 13. This is M-expr
;; syntax, obviously.
;; apply
;; NOTE THAT I suspect there is a typo in the printed manual in line
;; 7 of this definition, namely a missing closing square bracket before
;; the final semi-colon; that has been corrected here.
apply[fn;args;a] = [
null[fn] -> NIL;
atom[fn] -> [get[fn;EXPR] -> apply[expr; args; a];
get[fn;SUBR] -> {spread[args];
$ALIST := a;
TSX subr4, 4};
T -> apply[cdr[sassoc[fn; a; λ[[]; error[A2]]]]; args a]];
eq[car[fn]; LABEL] -> apply[caddr[fn]; args;
cons[cons[cadr[fn];caddr[fn]]; a]];
eq[car[fn]; FUNARG] -> apply[cadr[fn]; args; caddr[fn]];
eq[car[fn]; LAMBDA] -> eval[caddr[fn]; nconc[pair[cadr[fn]; args]; a]];
T -> apply[eval[fn;a]; args; a]]

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;; see page 70 of Lisp 1.5 Programmers Manual; this expands somewhat
;; on the accounts of eval and apply given on page 13. This is M-expr
;; syntax, obviously.
;; ## APPLY
;; NOTE THAT I suspect there is a typo in the printed manual in line
;; 7 of this definition, namely a missing closing square bracket before
;; the final semi-colon; that has been corrected here.
;; RIGHT! So the 'EXPR' representation of a function is expected to be
;; on the `EXPR` property on the property list of the symbol which is
;; its name; an expression is simply a Lisp S-Expression as a structure
;; of cons cells and atoms in memory. The 'SUBR' representation, expected
;; to be on the `SUBR` property, is literally a subroutine written in
;; assembly code, so what is happening in the curly braces is putting the
;; arguments into processor registers prior to a jump to subroutine - TSX
;; being presumably equivalent to a 6502's JSR call.
;; This accounts for the difference between this statement and the version
;; on page 12: that is a pure interpreter, which can only call those host
;; functions that are explicitly hard coded in.
;; This version knows how to recognise subroutines and jump to them, but I
;; think that by implication at least this version can only work if it is
;; itself compiled with the Lisp compiler, since the section in curly braces
;; appears to be intended to be passed to the Lisp assembler.
;; apply[fn;args;a] = [
;; null[fn] -> NIL;
;; atom[fn] -> [get[fn;EXPR] -> apply[expr; args; a];
;; get[fn;SUBR] -> {spread[args];
;; $ALIST := a;
;; TSX subr4, 4};
;; T -> apply[cdr[sassoc[fn; a; λ[[]; error[A2]]]]; args a]];
;; eq[car[fn]; LABEL] -> apply[caddr[fn]; args;
;; cons[cons[cadr[fn];caddr[fn]]; a]];
;; eq[car[fn]; FUNARG] -> apply[cadr[fn]; args; caddr[fn]];
;; eq[car[fn]; LAMBDA] -> eval[caddr[fn]; nconc[pair[cadr[fn]; args]; a]];
;; T -> apply[eval[fn;a]; args; a]]

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@ -1,7 +1,22 @@
;; Not present in Lisp 1.5(!)
;; Page 12 of the manual; this does NOT do what I expect a modern
;; ASSOC to do!
;; Modern ASSOC would be:
;; assoc[x; l] = [null[l] -> NIL;
;; and[consp[car[l]]; eq[caar[l]; x]] -> cdar[l];
;; T -> assoc[x; cdr[l]]]
;; In the Lisp 1.5 statement of ASSOC, there's no account of what should happen
;; if the key (here `x`) is not present on the association list `a`. It seems
;; inevitable that this causes an infinite run up the stack until it fails with
;; stack exhaustion. Consequently this may be right but I'm not implementing it!
;; assoc[x; a] = [equal[caar[a]; x] -> car[a];
;; T -> assoc[x; cdr[a]]]
;; Consequently, my solution is a hybrid. It returns the pair from the
;; association list, as the original does, but it traps the end of list
;; condition, as a modern solution would.
assoc[x; l] = [null[l] -> NIL;
and[consp[car[l]]; eq[caar[l]; x]] -> cdar[l];
and[consp[car[l]]; eq[caar[l]; x]] -> car[l];
T -> assoc[x; cdr[l]]]
;; (ASSOC 'C (PAIR '(A B C D E F) (RANGE 1 6)))

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;; page 12
pairlis[x;y;a] = [null[x] -> a;
T -> cons[cons[car[x]; car[y]];
pairlis[cdr[x]; cdr[y]; a]]]

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;; There are two different statements of SUBLIS and SUB2 in the manual, on
;; pages 12 and 61 respectively, although they are said to be semantically
;; equivalent; this is the version from page 12.
sub2[a; z] = [null[a] -> z;
eq[caar[a]; z] -> cdar[a];
T -> sub2[cdar[a]; z]]
sublis[a; y] = [atom[y] -> sub2[a; y];
T -> cons[]]

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;; page 11
subst[x; y; z] = [equal[y; z] -> x;
atom[z] -> z;
T -> cons[subst[x; y; car[z]]; subst[x; y; cdr[z]]]]