方法1：
call date_and_time(charecter_date(len=8),charecter_time(len=10))
需要日期可以用这种方法
方法2：
call cpu_time(real_time)
最简单的方法，但只适用于单线程，
如果是多线程运行，得到的结果是所以线程时间的和！
方法3：
call system_clock(int_time)
整形，单位是毫秒，多线程用这个方法
代码：
program main
implicit none
real(kind=4) :: t1,t2
integer :: int1,int2
integer :: ii,jj,kk
character(len=10) :: time1,time2
character(len=8) :: thedate
character(len=12) :: thetime
write(*,*) 'date_and_time:'
call date_and_time(thedate,time1)
do ii = 1,10000
do jj = 1,100000
kk = jj
enddo
enddo
call date_and_time(thedate,time2)
write(*,*) 'date: ',thedate
write(*,*) 'time1: ',time1
write(*,*) 'time2: ',time2
write(*,*) 'time2 is ',time2(1:2) // ':' // time2(3:4) // ':' // time2(5:10)
read(time1,*) t1
read(time2,*) t2
write(*,*) 'time:', t2-t1
write(*,*) ' '
write(*,*) 'cpu_time:'
call cpu_time(t1)
do ii = 1,10000
do jj = 1,100000
kk = jj
enddo
enddo
call cpu_time(t2)
write(*,*) 't1:',t1
write(*,*) 't2:',t2
write(*,*) 't2-t1=',t2-t1
write(*,*) ' '
write(*,*) 'system_clock:'
call system_clock(int1)
do ii = 1,10000
do jj = 1,100000
kk = jj
enddo
enddo
call system_clock(int2)
write(*,*) 'int1: ', int1
write(*,*) 'int2: ', int2
write(*,*) 'unit is ms'
write(*,*) '(int2-int1)/1000. is ' , (int2-int1)/1000.
end
结果：
date_and_time:
date: 20190111
time1: 112638.795
time2: 112642.070
time2 is 11:26:42.070
time: 3.27343750
cpu_time:
t1: 3.27646708
t2: 6.54442978
t2-t1= 3.26796269
system_clock:
int1: 120586558
int2: 120589826
unit is ms
(int2-int1)/1000. is 3.26799989