PROGRAM EIGHTEENUse records for todays date, tomorrows date, An array can be used to hold the days for each month of the year.
Jan to Dec = 31,28,31,30,31,30,31,31,30,31,30,31Remember to change the month or year as necessary.
program PROG18 (input,output); {date calculation program}
type date = record
day, month, year : integer;
end;
datename = array[1..12] of integer;
procedure update( var tomorrow : date; days_in_month : datename );
begin
tomorrow.day := tomorrow.day + 1; {increment day}
if tomorrow.day > days_in_month[tomorrow.month] then
begin
tomorrow.day := 1;
tomorrow.month := tomorrow.month + 1; {adjust month }
if tomorrow.month > 12 then {adjust year }
begin
tomorrow.month := 1;
tomorrow.year := tomorrow.year + 1
end
end
end;
var todays_date : date;
days : datename;
begin
days[1] := 31; days[2] := 28; days[3] := 31; days[4] := 30;
days[5] := 31; days[6] := 30; days[7] := 31; days[8] := 31;
days[9] := 30; days[10] := 31; days[11] := 30; days[12] := 31;
writeln('Enter todays date dd mm yy ');
readln( todays_date.day, todays_date.month, todays_date.year);
update( todays_date, days );
writeln('Tomorrows date will be ', todays_date.day,'-',
todays_date.month,'-',todays_date.year)
end.