使用openmp和私有子句的梯形规则集成

Question

我正在更改用于串行执行的代码，将其调整为并行执行(openmp)，但我得到的结果(pi值)非常接近。我在下面展示了这两种密码。

有什么问题吗？

program trap
use omp_lib 
implicit none
double precision::suma=0.d0 ! sum is a scalar
double precision:: h,x,lima,limb
integer::n,i, istart, iend, thread_num, total_threads=4, ppt
integer(kind=8):: tic, toc, rate
double precision:: time
double precision, dimension(4):: pi= 0.d0

call system_clock(count_rate = rate)
call system_clock(tic)

lima=0.0d0; limb=1.0d0; suma=0.0d0; n=10000000
h=(limb-lima)/n

suma=h*(f(lima)+f(limb))*0.5d0 !first and last points

ppt= n/total_threads
!$ call omp_set_num_threads(total_threads)

!$omp parallel private (istart, iend, thread_num, i)
  thread_num = omp_get_thread_num()
  !$ istart = thread_num*ppt +1
  !$ iend = min(thread_num*ppt + ppt, n-1)
do i=istart,iend ! this will control the loop in different images
  x=lima+i*h
  suma=suma+f(x) 
  pi(thread_num+1)=suma
enddo
!$omp end parallel

suma=sum(pi) 
suma=suma*h

print *,"The value of pi is= ",suma ! print once from the first image
!print*, 'pi=' , pi
call system_clock(toc)
time = real(toc-tic)/real(rate)
print*, 'Time ', time, 's'

contains

double precision function f(y)
double precision:: y
f=4.0d0/(1.0d0+y*y)
end function f

end program trap

!----------------------------------------------------------------------------------
program trap
implicit none
double precision::sum ! sum is a scalar
double precision:: h,x,lima,limb
integer::n,i
integer(kind=8):: tic, toc, rate
double precision:: time

call system_clock(count_rate = rate)
call system_clock(tic)

lima=0.0d0; limb=1.0d0; sum=0.0d0; n=10000000
h=(limb-lima)/n

sum=h*(f(lima)+f(limb))*0.5d0 !first and last points

do i=1,n-1 ! this will control the loop in different images
  x=lima+i*h
  sum=sum+f(x)
enddo

sum=sum*h

print *,"The value of pi is (serial exe)= ",sum ! print once from the first image

call system_clock(toc)
time = real(toc-tic)/real(rate)
print*, 'Time serial execution', time, 's'

contains

double precision function f(y)
double precision:: y
f=4.0d0/(1.0d0+y*y)
end function f

end program trap

汇编使用：

$ gfortran -fopenmp -Wall -Wextra -O2 -Wall -o prog.exe test.f90 
$ ./prog.exe

和

$ gfortran -Wall -Wextra -O2 -Wall -o prog.exe testserial.f90 
$ ./prog.exe

在串行执行中，我得到了pi (3.1415)的很好的近似，但是使用并行得到(我展示了几个并行执行)：

 The value of pi is=    3.6731101425922810     

 Time    3.3386986702680588E-002 s

-------------------------------------------------------

 The value of pi is=    3.1556004791445953     

 Time    8.3681479096412659E-002 s

------------------------------------------------------

 The value of pi is=    3.2505952856717966     

 Time    5.1473543047904968E-002 s

Answer 1

openmp并行语句中有一个问题。继续添加到变量suma上。因此，您需要指定一个reduction语句。此外，您也没有指定变量x为私有变量。

我还更改了代码的更多部分。

• ，您显式地告诉每个线程它应该使用哪个索引范围。大多数情况下，编译器可以自己更有效地解决这个问题。为此，我将parallel改为parallel do。
• 在openmp并行区域中将变量属性设置为default(none)是很好的做法。您需要显式地设置每个变量属性。

program trap
  use omp_lib
  implicit none
  double precision   :: suma,h,x,lima,limb, time
  integer            :: n, i
  integer, parameter :: total_threads=5
  integer(kind=8)    :: tic, toc, rate

  call system_clock(count_rate = rate)
  call system_clock(tic)

  lima=0.0d0; limb=1.0d0; suma=0.0d0; n=10000000
  h=(limb-lima)/n

  suma=h*(f(lima)+f(limb))*0.5d0 !first and last points

  call omp_set_num_threads(total_threads)
  !$omp parallel do default(none) private(i, x) shared(lima, h, n)  reduction(+: suma)
  do i = 1, n
    x=lima+i*h
    suma=suma+f(x)
  end do
  !$omp end parallel do

  suma=suma*h

  print *,"The value of pi is= ", suma ! print once from the first image
  call system_clock(toc)
  time = real(toc-tic)/real(rate)
  print*, 'Time ', time, 's'

contains

  double precision function f(y)
    double precision:: y
    f=4.0d0/(1.0d0+y*y)
  end function

end program