intelvisualfortran在visualstudio中如何正常的使用openmp并行程序文档格式.docx

1、 However, the parallel version computes the sum in a different order than the serial version; some of the quantities added are quite small, and so this will affect the accuracy of the results. Modified: 29 January 2003 Author: John Burkardt A FORTRAN 90 module may be available: use omp_lib A FORTRAN

2、 77 include file may be available: include omp_lib.h implicit none integer, parameter : r4_logn_max = 9 integer id integer nthreads integer omp_get_num_procs integer omp_get_num_threads integer omp_get_thread_num call timestamp （ ） write （ *, （a） ） TEST_OMP FORTRAN90 version Estimate the value of PI

3、 by summing a series. This program includes Open MP directives, which may be used to run the program in parallel. The number of processors available:（a,i8） OMP_GET_NUM_PROCS （） = , omp_get_num_procs （ ） nthreads = 4（a,i8,a） Call OMP_SET_NUM_THREADS, and request , & nthreads, threads. call omp_set_nu

4、m_threads （ nthreads ） Note that the call to OMP_GET_NUM_THREADS will always return 1 if called outside a parallel region!$OMP parallel private （ id ） id = omp_get_thread_num （ ）（a,i3） This is process , id if （ id = 0 ） then Calling OMP_GET_NUM_THREADS inside a parallel region, we get the number of

5、threads is , omp_get_num_threads （ ） end if$OMP end parallel call r4_test （ r4_logn_max ） Normal end of execution. stopendsubroutine r4_test （ logn_max ） R4_TEST estimates the value of PI using single precision. PI is estimated using N terms. N is increased from 102 to 10LOGN_MAX. The calculation is

6、 repeated using both sequential and Open MP enabled code. Wall clock time is measured by calling SYSTEM_CLOCK. 06 January 2003 integer clock_max integer clock_rate integer clock_start integer clock_stop real error real estimate integer logn integer logn_max character （ len = 3 ） mode integer n real

7、r4_pi real timeR4_TEST: Estimate the value of PI, using single precision arithmetic. N = number of terms computed and added; ESTIMATE = the computed estimate of PI; ERROR = （ the computed estimate - PI ）; TIME = elapsed wall clock time; Note that you cant increase N forever, because: A） ROUNDOFF sta

8、rts to be a problem, and B） maximum integer size is a problem.（a,i12） The maximum integer: , huge （ n ） N Mode Estimate Error Time n = 1 do logn = 2, logn_max mode = OMP call system_clock （ clock_start, clock_rate, clock_max ） call r4_pi_est_omp （ n, estimate ） call system_clock （ clock_stop, clock_

9、rate, clock_max ） time = real （ clock_stop - clock_start ） / real （ clock_rate ） error = abs （ estimate - r4_pi （ ） ）（ i12, 2x, a3, 2x, f14.10, 2x, g14.6, 2x, g14.6 ） ） & n, mode, estimate, error, time n = n * 10 end do returnsubroutine r4_pi_est_omp （ n, estimate ） R4_PI_EST_OMP estimates the value

10、 of PI, using Open MP. The calculation is based on the formula for the indefinite integral: Integral 1 / （ 1 + X*2 ） dx = Arctan （ X ） Hence, the definite integral Integral （ 0 = X = 1 ） 1 / （ 1 + X*2 ） dx = Arctan （ 1 ） - Arctan （ 0 ） = PI / 4. A standard way to approximate an integral uses the mid

11、point rule. If we create N equally spaced intervals of width 1/N, then the midpoint of the I-th interval is X（I） = （2*I-1）/（2*N）. The approximation for the integral is then: Sum （ 1 = I = N ） （1/N） * 1 / （ 1 + X（I）*2 ） In order to compute PI, we multiply this by 4; we also can pull out the factor of

12、 1/N, so that the formula you see in the program looks like: （ 4 / N ） * Sum （ 1 = N ） 1 / （ 1 + X（I）*2 ） Until roundoff becomes an issue, greater accuracy can be achieved by increasing the value of N. Parameters: Input, integer N, the number of terms to add up. Output, real ESTIMATE, the estimated

13、value of pi. real h integer i real sum2 real x h = 1.0E+00 / real （ 2 * n ） sum2 = 0.0E+00$OMP parallel do private（x） shared（h） reduction（+: sum2） do i = 1, n x = h * real （ 2 * i - 1 ） sum2 = sum2 + 1.0E+00 / （ 1.0E+00 + x*2 ） estimate = 4.0E+00 * sum2 / real （ n ）function r4_pi （ ） R4_PI returns t

14、he value of pi. 02 February 2000 Output, real R4_PI, the value of pi. r4_pi = 3.14159265358979323846264338327950288419716939937510E+00subroutine timestamp （ ） TIMESTAMP prints the current YMDHMS date as a time stamp. Example: May 31 2001 9:45:54.872 AM 31 May 2001 None character （ len = 8 ） ampm int

15、eger d character （ len = 8 ） date integer h integer m integer mm character （ len = 9 ）, parameter, dimension（12） : month = （/ &January , February March April May June July August SeptemberOctober November December /） integer s character （ len = 10 ） time integer values（8） integer y character （ len =

16、 5 ） zone call date_and_time （ date, time, zone, values ） y = values（1） m = values（2） d = values（3） h = values（5） n = values（6） s = values（7） mm = values（8） if （ h 12 ） then ampm = AM else if （ h = 12 ） then if （ n = 0 .and. s = 0 ） thenNoon elsePM h = h - 12Midnight（a,1x,i2,1x,i4,2x,i2,a1,i2.2,a1,i2.2,a1,i3.3,1x,a） trim （ month（m） ）, d, y, h, , n, , s, ., mm, trim （ ampm ）=COPY上面的程序，可以完全运行成功，运行界面如下：