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81 lines
2.2 KiB
Forth
81 lines
2.2 KiB
Forth
( requires core, parse )
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( Managing variables in a core module is tricky. Sure, we
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have (sysv), but here we need to allocate a big buffer, and
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that cannot be done through (sysv). What we do is that we
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allocate that buffer at runtime and use (sysv) to point to
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it, a pointer that is set during the initialization
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routine. )
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64 CONSTANT INBUFSZ
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( points to INBUF )
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(sysv) IN(
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( points to INBUF's end )
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(sysv) IN)
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( current position in INBUF )
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(sysv) IN>
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( flush input buffer )
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( set IN> to IN( and set IN> @ to null )
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: (infl) 0 IN( @ DUP IN> ! ! ;
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( Initializes the readln subsystem )
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: (c<$)
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H@ IN( !
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INBUFSZ ALLOT
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H@ IN) !
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( We need two extra bytes. 1 for the last typed 0x0a and
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one for the following NULL. )
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2 ALLOT
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(infl)
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;
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( handle backspace: go back one char in IN>, if possible, then
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emit SPC + BS )
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: (inbs)
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( already at IN( ? )
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IN> @ IN( @ = IF EXIT THEN
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IN> @ 1 - IN> !
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SPC BS
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;
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( read one char into input buffer and returns whether we
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should continue, that is, whether newline was not met. )
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: (rdlnc) ( -- f )
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( buffer overflow? same as if we typed a newline )
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IN> @ IN) @ = IF 0x0a ELSE KEY THEN ( c )
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( del? same as backspace )
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DUP 0x7f = IF DROP 0x8 THEN
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( echo back )
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DUP EMIT ( c )
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( bacspace? handle and exit )
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DUP 0x8 = IF (inbs) EXIT THEN
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( write and advance )
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DUP ( keep as result ) ( c c )
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( Here, we take advantage of the fact that c's MSB is
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always zero and thus ! automatically null-terminates
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our string )
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IN> @ ! 1 IN> +! ( c )
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( if newline, replace with zero to indicate EOL )
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DUP 0xa = IF DROP 0 THEN
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;
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( Read one line in input buffer and make IN> point to it )
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: (rdln)
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( Should we prompt? if we're executing a word, FLAGS bit
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0, then we shouldn't. )
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FLAGS @ 0x1 AND NOT IF LF '>' EMIT SPC THEN
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(infl)
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BEGIN (rdlnc) NOT UNTIL
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IN( @ IN> !
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;
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( And finally, implement a replacement for the (c<) routine )
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: (c<)
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IN> @ C@ ( c )
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( not EOL? good, inc and return )
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DUP IF 1 IN> +! EXIT THEN ( c )
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( EOL ? readline. we still return typed char though )
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(rdln) (<c) ( c )
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;
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