78 lines
1.9 KiB
NASM
78 lines
1.9 KiB
NASM
; *** Const ***
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; Base of the Return Stack
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.equ RS_ADDR 0xf000
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; Number of bytes we keep as a padding between HERE and the scratchpad
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.equ PADDING 0x20
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; Max length of dict entry names
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.equ NAMELEN 8
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; Offset of the code link relative to the beginning of the word
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.equ CODELINK_OFFSET 10
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; When set, the interpreter should abort parsing of current line and return to
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; prompt.
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.equ FLAG_QUITTING 0
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; When set, the interpreter should quit
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.equ FLAG_ENDPGM 1
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; *** Variables ***
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.equ INITIAL_SP FORTH_RAMSTART
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.equ CURRENT @+2
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.equ HERE @+2
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.equ INPUTPOS @+2
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.equ FLAGS @+2
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; Buffer where we compile the current input line. Same size as STDIO_BUFSIZE.
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.equ COMPBUF @+1
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.equ FORTH_RAMEND @+0x40
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; *** Code ***
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MAIN:
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.dw compiledWord
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.dw INTERPRET+CODELINK_OFFSET
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.dw CHKEND
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; If FLAG_ENDPGM is set, stop the program, else, tweak the RS so that we loop.
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CHKEND:
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.dw nativeWord
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ld hl, FLAGS
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bit FLAG_ENDPGM, (hl)
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jr nz, .endpgm
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; not quitting program, are we supposed to continue parsing line?
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ld hl, FLAGS
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bit FLAG_QUITTING, (hl)
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jr nz, forthRdLine
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; Not quitting line either.
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jr forthInterpret
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.endpgm:
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ld sp, (INITIAL_SP)
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; restore stack
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pop af \ pop af \ pop af
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xor a
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ret
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forthMain:
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; STACK OVERFLOW PROTECTION:
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; To avoid having to check for stack underflow after each pop operation
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; (which can end up being prohibitive in terms of costs), we give
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; ourselves a nice 6 bytes buffer. 6 bytes because we seldom have words
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; requiring more than 3 items from the stack. Then, at each "exit" call
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; we check for stack underflow.
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push af \ push af \ push af
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ld (INITIAL_SP), sp
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ld hl, CONSTANT ; last entry in hardcoded dict
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ld (CURRENT), hl
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ld hl, FORTH_RAMEND
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ld (HERE), hl
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forthRdLine:
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xor a
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ld (FLAGS), a
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ld hl, msgOk
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call printstr
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call printcrlf
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call stdioReadLine
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ld (INPUTPOS), hl
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forthInterpret:
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ld ix, RS_ADDR
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ld iy, MAIN
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jp executeCodeLink
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msgOk:
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.db " ok", 0
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