[texhax] Time calculations within LaTeX?

[E. Krishnan]

On Sat, 9 Feb 2008, Richard Hartmann wrote:

> On Feb 9, 2008 6:56 AM, E. Krishnan <ekmath at asianetindia.com> wrote:
>
>
>> Then \timediff10:15-16:00 gives 5.75 and \timediff10:15-01:00 gives 14.75.
>
> I found two issues with the function, one of which I could fix :)
>
> You need to reinitialize dh, dm, hrdec for the function to give the correct
> result when called more than once in a tex file.
>
> \setcounter{dh}{0}
> \setcounter{dm}{0}
> \setcounter{hrdec}{0}
>
> does the trick.


But I didn't have this issue.


> The other issue is that when the minute or hour values are equal, you
> get a null result, no matter what the other parameter is.


Oops! Sort of overlooked that case. Here's the modification to take care 
of this:

  \def\timediff#1:#2-#3:#4 {%
   \setcounter{bh}{#1}
   \setcounter{bm}{#2}
   \setcounter{eh}{#3}
   \setcounter{em}{#4}
   \ifthenelse{\(\value{eh}>\value{bh}\or\value{eh}=\value{bh}\)
               \and\(\value{em}>\value{bm}\or\value{em}=\value{bm}\)}
     {\setcounter{dh}{\value{eh}-\value{bh}}
      \setcounter{dm}{\value{em}-\value{bm}}}
     {\ifthenelse{\(\value{eh}>\value{bh}\or\value{eh}=\value{bh}\)
                  \and\value{em}<\value{bm}}
       {\setcounter{dh}{\value{eh}-\value{bh}-1}
        \setcounter{dm}{60+\value{em}-\value{bm}}}
       {\ifthenelse{\value{eh}<\value{bh}
         \and\(\value{em}>\value{bm}\or\value{em}=\value{bm}\)}
        {\setcounter{dh}{24+\value{eh}-\value{bh}}
         \setcounter{dm}{\value{em}-\value{bm}}}
        {\ifthenelse{\value{eh}<\value{bh}\and\value{em}<\value{bm}}
        {\setcounter{dh}{23+\value{eh}-\value{bh}}
         \setcounter{dm}{60+\value{em}-\value{bm}}}}}{}}
    \setcounter{hrdec}{\value{dm}*5/3}
    \thedh.\thehrdec}


-- 
Krishnan