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Double( 6)' would be an incorrect call of this procedure, since 6 is not a name. An example more relevant to this chapter would be a procedure to advance the date to the next working day, assuming a 5-day week and ignoring public holidays. 'proc advance date=( char day, ref int date,month,year)void: (int n = nodays in(month, year); date plusab (day= "FRI" 131 : day= "SAT" 12 II); C 1: is an abbreviation of elif C if date> n then date rninusab n; if month= I2 then month := I ; year plusab I else month plusab I fi fi)' The parameters 'month', 'year' and 'date' are all ref int because they may be altered by the procedure.
Later we shall discuss 'format' for the attractive arrangement of output. This is the output of the above program for a particular maze SQUARE MAZE OF SIZE . *. * ENTRANCE COORDS = I 3 3 EXIT COORDS = 3 2 MAZE SOLVED BY FOLLOWING THE PATH WWSSE When developing the program, additional information should be printed to help debugging, such as the coordinates of every move made and of the dead ends pruned. The special 'print' instructions can be removed when the program works correctly. ) TRAVERSING A MAZE: MULTIPLE VALUES 35 The algorithm should also be tested on data where there is no route out, on illegal data and on a maze where several dead ends are encountered before a path is found.
1. : 1973. 2. Calculate advance days(y). 3. Calculate the number b of days from 1 Jan y to the given date, noting that if y is a leap year, Feb has 29 days. 4. 1 Jan 1973 was a Monday, code number 1, so print the day whose code is (1 +advance days(y) +b) mod 7. 1 Perform this algorithm for the date 29 Feb 2000. Apart from the input and output, the algorithm comprises two logically distinct calculations, and their isolation as separate steps makes it easier to follow and check. A self-contained piece of program that takes some data like 'y', or even no data, and defines some operation possibly producing a result, like 'advance days(y)', can be made into a procedure.