beowulf/resources/mexpr/apply.mexpr.lsp

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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]]