Euphoria 程序庫例程返回的日期和時間。

date() 方法:

date() 方法返回八個原子元素組成的序列值。以下舉例說明它的細節:


integer curr_year, curr_day, curr_day_of_year, curr_hour,
curr_minute, curr_second
sequence system_date, word_week, word_month, notation,
curr_day_of_week, curr_month
word_week = {"Sunday",
word_month = {"January", "February",
"March", "April", "May",
"June", "July", "August",
"September", "October",
"November", "December"}
-- Get current system date.
system_date = date()

-- Now take individual elements
curr_year = system_date[1] + 1900
curr_month = word_month[system_date[2]]
curr_day = system_date[3]
curr_hour = system_date[4]
curr_minute = system_date[5]
curr_second = system_date[6]
curr_day_of_week = word_week[system_date[7]]
curr_day_of_year = system_date[8]

if curr_hour >= 12 then
notation = "p.m."
notation = "a.m."
end if

if curr_hour > 12 then
curr_hour = curr_hour - 12
end if
if curr_hour = 0 then
curr_hour = 12
end if

puts(1, "\nHello Euphoria!\n\n")
printf(1, "Today is %s, %s %d, %d.\n",
{curr_day_of_week, curr_month,
curr_day, curr_year})

printf(1, "The time is %.2d:%.2d:%.2d %s\n",
{curr_hour, curr_minute,
curr_second, notation})

printf(1, "It is %3d days into the current year.\n",


Hello Euphoria!

Today is Friday, January 22, 2010.
The time is 02:54:58 p.m.
It is 22 days into the current year.

time() 方法:

The time() 方法返回一個原子值,相當於一個固定的時間點以來經過的秒數。以下舉例說明它的細節:


constant ITERATIONS = 100000000
integer p
atom t0, t1, loop_overhead

t0 = time()
for i = 1 to ITERATIONS do
-- time an empty loop
end for

loop_overhead = time() - t0

printf(1, "Loop overhead:%d\n", loop_overhead)

t0 = time()
for i = 1 to ITERATIONS do
p = power(2, 20)
end for

t1 = (time() - (t0 + loop_overhead))/ITERATIONS

printf(1, "Time (in seconds) for one call to power:%d\n", t1)


Loop overhead:1
Time (in seconds) for one call to power:0

Date & Time 相關方法:

Euphoria 提供了許多方法,它可以幫助操縱日期和時間。前方法中列出 Euphoria Library Routines.