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[求助]
Fluent UDF 改單機(jī)多核
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最近在學(xué)習(xí)udf的并行，也看了很多資料，仍然不能掌握，所以給出一個(gè)例子請(qǐng)求幫助。 以下是一個(gè)簡單的定義物體x方向上的速度每一個(gè)時(shí)間步后速度增加0.001，如何改成單機(jī)多核，或者說不用改就能在單機(jī)多核上運(yùn)行？ 謝謝！ #include "udf.h" #include "mem.h" #include "math.h" #include "dynamesh_tools.h" static real init_vel=0; DEFINE_CG_MOTION(sway,dt,vel,omega,time,dtime) { FILE *fd; init_vel+=0.001;/*迭代出新的速度*/ vel[0]=init_vel; fd=fopen("vel.dat","a"  ;fprintf(fd,"%f,%f\n", time, vel[0]); fclose(fd); } |
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補(bǔ)充一點(diǎn)： 我的電腦是單機(jī)4核，用的15.0，采用的串行，編譯上述udf得到的速度更新結(jié)果如下：它并不是每一個(gè)時(shí)間步速度增加，而是每個(gè)時(shí)間步內(nèi)速度已經(jīng)疊加16次，并且我也嘗試過和每一步的迭代次數(shù)無關(guān)。 不知道問題出在哪里了，請(qǐng)有經(jīng)驗(yàn)的朋友指點(diǎn)！感謝！ timestep 速度 0.001000,0.001 0.001000,0.002 0.001000,0.003 0.001000,0.004 0.001000,0.005 0.001000,0.006 0.001000,0.007 0.001000,0.008 0.001000,0.009 0.001000,0.010 0.001000,0.011 0.001000,0.012 0.001000,0.013 0.001000,0.014 0.001000,0.015 0.002000,0.016 0.002000,0.017 0.002000,0.018 0.002000,0.019 0.002000,0.020 0.002000,0.021 0.002000,0.022 0.002000,0.023 0.002000,0.024 0.002000,0.025 0.002000,0.026 0.002000,0.027 0.002000,0.028 0.002000,0.029 0.002000,0.030 0.002000,0.031 0.002000,0.032 |
2樓2015-11-14 17:23:16
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