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Odd word problem: Difference between revisions
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No recursion or the stack is used. The program could've been written without subroutines.
<lang rexx>/*REXX program solves the odd word problem by only using byte input/output. */
iFID_ = 'ODDWORD.IN' /*Note: numeric suffix is added later.*/
oFID_ = 'ODDWORD.' /* " " " " " " */
do case=1 for 2; #=0 /*#: is the number of characters read.*/
iFID=iFID_ || case /*read ODDWORD.IN1 or ODDWORD.IN2 */
oFID=oFID_ || case /*write ODDWORD.1 o r ODDWORD.2 */
say; say; say '════════ reading file: ' iFID "════════"
do until x==. /* [↓] perform DO loop for odd words.*/
do until \isMix(x); call readChar; call writeChar
if x==. then leave /*is this end─of─sentence? (full stop)*/
call readLetters; punctuation_location=#
do j=#-1 by -1; call readChar j
if \isMix(x) then leave; call writeChar
end /*j*/ /* [↑] perform for the "even" words.*/
call readLetters; call writeChar; #=punctuation_location
end /*until x ···*/
end /*case*/ /* [↑] process both of the input files*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isMix: return datatype(arg(1),'M') /*return 1 if
readLetters: do until \isMix(x); call readChar; end; return
writeChar: call charout ,x; call charout oFID,x; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
else x=charin(iFID, arg(1)) /* " specific "*/
return</lang>
'''output''' when using two (default) input files which contain:
:* input file '''ODDWORD.IN1''' ───► <tt> what,is,the;meaning,of:life. </tt>
:* input file '''ODDWORD.IN2''' ───► <tt> we,are;not,in,kansas;any,more. </tt>
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