Language RPAL
Date: | 05/29/05 |
Author: | Manuel E. Bermudez |
URL: | http://www.cise.ufl.edu/~manuel |
Comments: | 1 |
Info: | n/a |
Score: | (3.01 in 103 votes) |
let s1 f = 'ottle' @Conc (f eq 1 -> 's' | '') @Conc ' of beer' // s1 returns one of two strings, depending on f: // f=1: 'ottles of beer' // f=*: 'ottle of beer' (*) anything but in let s2 = ' on the wall' in let s3 f1 f2 = (f1 eq 0 -> ' B' | ' b') @Conc s1 f2 @Conc s2 @Conc (f1 eq 0 -> ', ' | '.\n\n') // s3 returns one of four strings, depending on f1, f2: // f1=0, f2=0: 'Bottle of beer on the wall, ' // f1=0, f2=1: 'Bottles of beer on the wall, ' // f1=*, f2=0: 'bottle of beer on the wall.\n\n ' // f1=*, f2=*: 'bottles of beer on the wall.\n\n ' in let s4 = 'Take one down and pass it around, ' in let s5 = 'Go to the store and buy some more...99 ' in let s6 = 'No more ' in let rec bottles n = n eq 0 -> // LAST LINE (case n=0): s6 @Conc 'b' // 'No more b' @Conc s1 0 // 'ottle of beer' @Conc s2 // ' on the wall' @Conc '. ' // '.' @Conc s6 // 'No more' @Conc 'b' // 'b' @Conc s1 1 // 'ottles of beer' @Conc '...\n' // '...\n' @Conc s5 // 'Go to the store ... 99' @Conc 'b' // 'b' @Conc s1 1 // 'ottles of beer' @Conc '.' // '.' // LINE n: 0 <= n <= 99 | ItoS n // n @Conc s3 0 (n eq 1 -> 0 | 1) // 'Bottle(s) of beer on the wall,' @Conc ItoS n // n @Conc ' b' // b @Conc s1 (n eq 1 -> 0 | 1) // ottle(s) of beer @Conc '.\n' // .\n @Conc s4 // Take one down ... , @Conc (n eq 1 -> 'no more' | ItoS (n-1)) // No more (or n-1) @Conc s3 1 (n eq 1 -> 1 | n > 2 -> 1 | 0) // bottle(s) of beer on the wall.\n\n @Conc bottles (n-1) // RECURSIVE CALL in Print (bottles 99)
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Comments
Daniel Franke said on 01/30/06 15:49:30
Very nice. I'm adding this as a test case in the interpreter distribution.