Comma quibbling: Difference between revisions

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=={{header|dc}}==

<syntaxhighlight lang="dc">[main entry point: print brackets, ~~call~~ o if stack not ~~empty]sx~~

<syntaxhighlight lang="dc">[(q)uibble: main entry point. print brackets, calling n in between if stack not

empty.]sx

[ [{]n z 0 !=o [}]pR ]sq

[ [{]n z 0 !=n [}]pR ]sq

[~~nonempty stack~~: if more than 1 item, call t. then print top of stack.]sx

[(n)onempty: if more than 1 item, call m. then print top of stack.]sx

[ z 1 !=t n ]so

[ z 1 !=m n ]sn

[~~more~~ ~~than~~ 1 ~~item.~~ ~~reverse (~~flip) stack into r register, call p to print

[(m)ore: call f to flip stack into r register, call p to print most of it,

~~most of it,~~ then pop the last item back ~~from~~ r so it's there to be printed

then pop the last item back onto the main stack so it's there to be printed

after we return]sx

[ lfx lpx Lr ]st

[ lfx lpx Lr ]sm

[utility routine: ~~reverse~~ ~~(flip)~~ the stack into the r register]sx

[(f)lip: utility routine to reverse the stack into the r register]sx

[ Sr z 0 !=f ]sf

[~~main print routine.~~ get next item from stack in r register,

[(p)rint: get next item from stack in r register and print it. If there are

~~print it. If there are~~ more than 2 items left, print a comma (c). ~~Otherwise,~~

more than 2 items left, print a comma (c) and recurse. If there are exactly

~~if there are exactly~~ two items left, print " and " (a). ~~Finally,~~ ~~if there~~

two items left, print " and " (a) and return.]sx

are more than 2 items left, recurse.]sx

[ Lr n 2 yr >c 2 yr =a 2 yr >p]sp

[utility routine: print a comma followed by a space]sx

[(c)omma: utility routine to print a comma followed by a space]sx

[ [, ]n ]sc

[utility routine: ~~print~~ a " and "]sx

[(a)and: utility routine to print " and "]sx

[ [ and ]n ]sa

[run tests]sx

lqx

[ABC] lqx

[ABC] [DEF] lqx

[ABC] [DEF] [G] [H] lqx

[ABC] [DEF] [G] [H] lqx</syntaxhighlight>

</syntaxhighlight>