```Contributor: LEE BARKER

{
"Does anyone have a decent routine for converting a
date to and from the number of days since 1/1/1970,
which properly takes into account leap years?"

While you can use Julian math, I use/wrote the following-
(Note: An integer can hold up to a little over 89 years,
or a word can hold upto 65536 days or about 179 years)
}

function leapyear (c,y : byte) : boolean;
begin
if (y and 3) <> 0
then leapyear := false
else if y=0
then leapyear := (c and 3)=0
else leapyear := true;
end;

function DaysInMonth (c,y,m : byte) : integer;
begin
if m=2
then if leapyear(c,y)
then DaysInMonth := 29
else DaysInMonth := 28
else DaysInMonth := 30 + ((\$15AA shr m) and 1);
end;

function DaysInYear (c,y : byte) : integer;
begin
DaysInYear := DaysInMonth(c,y,2)+337;
end;

Function DayOfYear (c,y,m,d :byte) : integer;
var i,j : integer;
begin
j := d;
for i := 1 to pred(m) do j := j + DaysInMonth(c,y,i);
DayOfYear := j;
end;

So for date2-date1
x := DaysInYear(date1) - DatOfYear(Date1);
for i := succ(date1) to pred(date2) do
x := x + DaysInYear(i);
x := x + DayOfYear(date2);
```