is_leap_year() #@ Return successfully if YEAR (or current year) is a leap year
{ #@ USAGE: is_leap_year [YEAR]
local year=$1 gregorian=1752
[[ -z $year ]] && printf -v year '%(%Y)T'
if (( year > gregorian ))
then
case $year in
*0[48] |\
*[2468][048] |\
*[13579][26] |\
*[02468][048]00 |\
*[13579][26]00 ) return 0 ;;
*) return 1
esac
else
(( year % 4 == 0 ))
fi
}