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标题: 一个循环嵌套的设计问题. [打印本页]
作者: 再见大麋鹿    时间: 2016-12-20 01:42
标题: 一个循环嵌套的设计问题.
本帖最后由 再见大麋鹿 于 2016-12-20 01:51 编辑

鄙人 Fortran 小白, 此问题查阅大量资料后无果, 且时间紧张, 故前来请教.
我的程序的思路是:
做两个自变量: time, voltage 的循环
就是先对 time 做一个循环，然后在每一个 time 数值做一个 voltage 的循环
最后得出2个自变量的函数 current=current(voltage, time)

而且我在给定 voltage 的时候是成功的, 也就是说, 原本只有一个 time 的循环, 现在我要在 time 的循环下加上一个 voltage 的循环.

原代码循环部分如下:

[Fortran] 纯文本查看 复制代码
time = 0.0d0

do

if (time .ge. time_end) exit



if (time <= time_center) then

    e_rabi_0 = e_rabi_i*exp(-(time-time_center)**2/time_width**2)       ! = exp(-4*(time-time_center)**2/time_center**2)

else

    e_rabi_0 = e_rabi_i

end if

    e_rabi_1 = e_rabi_0



call exci_rates(de,sq_fcf,gam,kop,e_rabi_0,e_rabi_1)

do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

do nc = 0, 1

do ne = 0, 1

do nv = 0, mv_max

    k_ref(mc,me,mv,nc,ne,nv) = kivr(mc,me,mv,nc,ne,nv) + kop(mc,me,mv,nc,ne,nv) + klu(mc,me,mv,nc,ne,nv)

end do

end do

end do

end do

end do

end do



voltage = voltage_give 





call mol_lead_rates(voltage,de,v_l,v_r,sq_fcf,klw,klw_l)

do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

do nc = 0, 1

do ne = 0, 1

do nv = 0, mv_max

    k_tot(mc,me,mv,nc,ne,nv) = k_ref(mc,me,mv,nc,ne,nv) + klw(mc,me,mv,nc,ne,nv)

end do

end do

end do

end do

end do

end do



if (time == 0.0d0) then

do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

    pop(mc,me,mv) = 0.d0

end do

end do

end do

pop(0,0,0) = 1.d0



stop_time = time_final

100 do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

    pop_old(mc,me,mv) = pop(mc,me,mv)

end do

end do

end do

call multi_to_single(pop_old,fin)

call rk4(fin,time,time_step,fout,k_tot)

call single_to_multi(fout,pop)



time = time + time_step

if (time < stop_time) goto 100



error_pop = 0.d0

do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

    error_pop = error_pop + (pop(mc,me,mv) - pop_old(mc,me,mv))**2

end do

end do

end do

    error_pop = SQRT(error_pop)/length_coeff

if (error_pop > max_error) then

    stop_time = stop_time + 0.5d0*time_final

end if

if (error_pop > max_error) goto 100

    time = 0.0d0

    time = time + time_step

else

do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

    pop_old(mc,me,mv) = pop(mc,me,mv)

end do

end do

end do



call multi_to_single(pop_old,fin)

call rk4(fin,time,time_step,fout,k_tot)

call single_to_multi(fout,pop) 

time = time + time_step

end if 



total_pop = 0.d0

mean_vib = 0.d0

do mc = 0, 1

do me = 0, 1

    pop_el(mc,me) = 0.d0

end do

end do



do mv = 0, mv_max

    pop_vib(mv) = 0.d0

end do

do mc = 0, 1

do me = 0, 1

do mv = 0, mv_max

    total_pop = total_pop + pop(mc,me,mv)

    pop_el(mc,me) = pop_el(mc,me) + pop(mc,me,mv)

end do

end do

end do



mean_vib=0.0d0

do mc = 0, 1     

do me = 0, 1   

do mv = 0, mv_max

    mean_vib = mean_vib + mv*pop(mc,me,mv)

end do

end do

end do



i_photon = 0.d0

do mc = 0, 1

do mv = 0, mv_max

do nv = 0, mv_max

    i_photon = i_photon + klu(mc,1,mv,mc,0,nv)*pop(mc,1,mv)

end do

end do

end do



current = 0.0d0

do me = 0, 1

do mv = 0, mv_max

do ne = 0, 1

do nv = 0, mv_max

    current = current + klw_l(0,me,mv,1,ne,nv)*pop(0,me,mv) - klw_l(1,ne,nv,0,me,mv)*pop(1,ne,nv)

end do

end do

end do

end do



do mc = 0, 1

do me = 0, 1

do nc = 0, 1

do ne = 0, 1

    c_in(mc,me,nc,ne) = 0.0d0

    c_out(mc,me,nc,ne) = 0.0d0

end do

end do

end do

end do



do mv = 0, mv_max

do nv = 0, mv_max

    c_in(0,0,1,0) = c_in(0,0,1,0) + klw_l(0,0,mv,1,0,nv)*pop(0,0,mv)

    c_in(0,0,1,1) = c_in(0,0,1,1) + klw_l(0,0,mv,1,1,nv)*pop(0,0,mv)

    c_in(0,1,1,0) = c_in(0,1,1,0) + klw_l(0,1,mv,1,0,nv)*pop(0,1,mv)

    c_in(0,1,1,1) = c_in(0,1,1,1) + klw_l(0,1,mv,1,1,nv)*pop(0,1,mv)

    

    c_out(1,0,0,0) = c_out(1,0,0,0) - klw_l(1,0,nv,0,0,mv)*pop(1,0,nv)

    c_out(1,0,0,1) = c_out(1,0,0,1) - klw_l(1,0,nv,0,1,mv)*pop(1,0,nv)

    c_out(1,1,0,0) = c_out(1,1,0,0) - klw_l(1,1,nv,0,0,mv)*pop(1,1,nv)

    c_out(1,1,0,1) = c_out(1,1,0,1) - klw_l(1,1,nv,0,1,mv)*pop(1,1,nv)

end do

end do



!!!!    Output

write (*,*) time, voltage, current*243.4d0

write (40,*) time, voltage, current*243.4d0



end do


这时候代码是可以正确运行的, 但是我想取若干个:
[Fortran] 纯文本查看 复制代码
voltage = voltage_give

所以我把这行改成了:
voltage = voltage_min
然后 Output 的前一行加上:
[Fortran] 纯文本查看 复制代码
if (num_voltage > 0) then      voltage = voltage + (voltage_max - voltage_min)/num_voltageend if

再然后, 末尾再加一个:
[Fortran] 纯文本查看 复制代码
end do

编译没错, 但是, 但是并不是每个 time 数值都做一个voltage的循环, 而是同时作循环:
[Fortran] 纯文本查看 复制代码
  1.000000000000000E-002   40.0000000000000       1.988160357654218E-016

  2.000000000000000E-002   80.0000000000000       1.364425784870382E-004

  3.000000000000000E-002   120.000000000000       1.370837425057141E-004

  4.000000000000000E-002   160.000000000000       2.083606552928942E-004

  5.000000000000000E-002   200.000000000000       3.832251385343335E-004

  6.000000000000000E-002   240.000000000000       3.846367630381155E-004

  7.000000000000001E-002   280.000000000000       4.131201522155061E-004

  8.000000000000000E-002   320.000000000000       5.825466767173463E-004

  9.000000000000000E-002   360.000000000000       5.852940467602035E-004

  0.100000000000000        400.000000000000       5.948948519956353E-004

  0.110000000000000        440.000000000000       7.259287371840456E-004

  0.120000000000000        480.000000000000       7.304897383455301E-004

……  

   38.1800000000010        2000.00000000000        74.8222446587716     

9


按理说应该是第一列先都是 1.000000000000000E-002 第二列从 00.0000000000000 每隔 40 数到 2000.000000000000   
然后第一列变成 2.000000000000000E-002 第二列重新从 00.0000000000000 每隔 40 数到 2000.000000000000   
……


近期是论文dead_line, 心态简直好到爆炸! T.T

最后感谢大家的帮助！


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作者: Jackdaw    时间: 2017-2-16 12:22
你可以吧这部分voltage 循环改成子程序，然后让这个子程序对time进行循环调用，输出结果，time作为参数每次传入一个数值



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