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[求助]
Fluent UDF 改單機多核
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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，所以給出一個例子請求幫助。 以下是一個簡單的定義物體x方向上的速度每一個時間步后速度增加0.001，如何改成單機多核，或者說不用改就能在單機多核上運行？ 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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補充一點： 我的電腦是單機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：它并不是每一個時間步速度增加，而是每個時間步內(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，請有經(jīng)驗的朋友指點！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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