date & time difference calculation

[Chris Sharman]

Lorry Litman wrote:
> Because I'm not familiar with these functions (or the math) in Chris's
> solution
> I have changed the imins define. 

> From:
> def imins int*2 = floor((fdays - idays)*1440 - 24 * ihours)
> To:
> def imins int*2 = floor((((fdays - idays)*24) - ihours)* 60 + .01)

Looks better (I'd incorrectly multiplied hours by 24 rather than 60) - 
well spotted. The 0.01 may help rounding issues, but will also introduce 
the possibility of imins being 60, which is probably undesirable.

Another issue is that floor rounds down, so -4.5 gets rounded to -5.
Not what we want at all.
Easiest fix:
define fdays = absolute(decimaltime(vmsdt1,vmsdt2))
...
define date_diff = ... if (vmsdt1>=vmsdt2) else - ...

Don't know whether that's the problem or not, but the following seems 
to give good results on a couple of test cases.

def fdays = absolute(decimaltime(vmsdt1,vmsdt2))
def idays int*4 = floor(fdays)
def ihours int*2 = floor((fdays - idays)*24)
def imins int*2 = floor(((fdays - idays)*24 - ihours)*60)
def idiff int*9 = idays*10000+ihours*100+imins if (vmsdt1>=vmsdt2) &
	else -(idays*10000+ihours*100+imins)
rep all
go

Chris



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