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Euphoria日期/时间

Euphoria 程序库例程返回的日期和时间。

date() 方法:

date() 方法返回八个原子元素组成的序列值。以下举例说明它的细节:

#!/home/euphoria-4.0b2/bin/eui
 
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", 
                     "Monday", 
                     "Tuesday", 
                     "Wednesday", 
                     "Thursday", 
                     "Friday", 
                     "Saturday"}
        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."
else 
   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", 
          {curr_day_of_year})

标准屏幕上,这将产生以下结果:

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() 方法返回一个原子值,相当于一个固定的时间点以来经过的秒数。以下举例说明它的细节:

#!/home/euphoria-4.0b2/bin/eui
 
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)

这将产生以下结果:

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

Date & Time 相关方法:

Euphoria 提供了许多方法,它可以帮助操纵日期和时间。前方法中列出 Euphoria Library Routines.