Fortran代码行依赖分析

1）先上代码

    !$omp parallel private(n, j, dx, dy, dz, r, a)
    do n = 1, nsteps
        !$omp do
        do i = 0, nparticles - 1
            x_tmp(i) = x(i) + vx(i) * dt
            y_tmp(i) = y(i) + vy(i) * dt
            z_tmp(i) = z(i) + vz(i) * dt

            do j = 0, nparticles - 1
                dx = x(j) - x(i)
                dy = y(j) - y(i)
                dz = z(j) - z(i)

                R = sqrt(dx ** 2 + dy ** 2 + dz ** 2 + softening)
                a = G * m(j) / R ** 3 * dt

                vx(i) = vx(i) + a * dx
                vy(i) = vy(i) + a * dy
                vz(i) = vz(i) + a * dz
            end do
        end do

        !$omp do
        do i = 0, nparticles - 1
            x(i) = x_tmp(i)
            y(i) = y_tmp(i)
            z(i) = z_tmp(i)
        end do

        !$omp master
        print *, "n =", n
        do i = 0, nparticles - 1
            print *, i, ":", x(i), y(i), z(i)
        end do
        !$omp end master
    end do
    !$omp end parallel

2）运行结果

4 <-- [24, 16]
5 <-- [25, 17]
6 <-- [26, 18]
9 <-- [24]
10 <-- [25]
11 <-- [26]
13 <-- [9, 10, 11]
14 <-- [13]
16 <-- [14, 9]
17 <-- [14, 10]
18 <-- [14, 11]
24 <-- [4]
25 <-- [5]
26 <-- [6]

3）分析

看起来是对的。