并行时函数只能在processor0执行
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Hello,各位大神，
我在用我自己写的一个求解器并行计算时出现了一个奇怪的问题。这个问题只有在并行计算的时候才会出来，当不并行时没事。问题的具体描述如下：
在我的solver中有一个名为 updateVelocity 的成员函数来获取计算域中某观察点（这个观察点的位置可能会随着时间改变）的速度。
具体如下:while (runTime.loop()) { Info << "Time = " << runTime.timeName() << nl << endl; #include "CourantNo.H" // Pressure-velocity PISO corrector { #include "UEqn.H" // --- PISO loop while (piso.correct()) { #include "pEqn.H" } } laminarTransport.correct(); turbulence->correct(); //>>>>>>>>>>>>>Below>>>>>>> Nettings.updateVelocity(U,mesh); //<<<<<<<<<<<<<Above<<<<<< runTime.write(); Info << "ExecutionTime = " << runTime.elapsedCpuTime() << " s" << " ClockTime = " << runTime.elapsedClockTime() << " s" << nl << endl; } Info << "End\n" << endl; return 0; }这个函数的源代码如下，具体实现形式就是判断每个cell的中心到观察点的距离，然后取出距离观察点距离最近的cell的速度
void Foam::netPanel::updateVelocity( const volVectorField &U, const fvMesh &mesh) { const vectorField ¢res(mesh.C()); List<vector> fluidVelocities(structuralElements_memb.size(), vector::zero); Info << "In updateVelocity, number of mesh is " << centres.size() << endl; Info << "In updateVelocity, number of U is " << U.size() << endl; scalar maxDistance(1); forAll(EPcenter, Elemi) { maxDistance = 1; vector nearestCell(vector::zero); scalar loops(0); forAll(centres, cellI) // loop through all the cells, { scalar k1(calcDist(centres[cellI], EPcenter[Elemi])); if (k1 < maxDistance) { maxDistance = k1; fluidVelocities[Elemi] = U[cellI]; nearestCell = centres[cellI]; loops += 1; Info << "After " << loops << " times of loop, the nearest cell is " << nearestCell << "to point " << EPcenter << "\n" << endl; } } } fluidVelocity_memb = fluidVelocities; // only assige onece Info << "the velocity on nodes are " << fluidVelocity_memb << endl; }就是这样的一个简单的函数，但是在并行时却出现了问题。
首先我们来看串行计算的情况：Starting time loop Time = 0.01 Courant Number mean: 0.000565 max: 0.0904 smoothSolver: Solving for Ux, Initial residual = 1, Final residual = 2.4007e-06, No Iterations 1 smoothSolver: Solving for Uy, Initial residual = 0.891308, Final residual = 1.23902e-06, No Iterations 1 smoothSolver: Solving for Uz, Initial residual = 0.895257, Final residual = 1.31102e-06, No Iterations 1 GAMG: Solving for p, Initial residual = 1, Final residual = 8.40918e-07, No Iterations 35 time step continuity errors : sum local = 9.50237e-10, global = 1.27782e-10, cumulative = 1.27782e-10 smoothSolver: Solving for epsilon, Initial residual = 1, Final residual = 0.00445555, No Iterations 1 smoothSolver: Solving for k, Initial residual = 1, Final residual = 0.004493, No Iterations 1 In updateVelocity, number of mesh is 184320 In updateVelocity, number of U is 184320 After 1 times of loop, the nearest cell is (-0.49375 -0.21875 -0.39375)to point (0 0.05 -0.1) ... ... After 45 times of loop, the nearest cell is (-0.00625 0.05625 -0.20625)to point (0 0.05 -0.2) the velocity on nodes are 4((0.226059 -2.8946e-08 -3.59708e-08) (0.226059 3.97379e-08 -4.07855e-08) (0.226059 2.65165e-08 -2.22689e-08) (0.226059 -4.12251e-08 -1.8172e-08)) ExecutionTime = 2.31 s ClockTime = 2 s计算的结果没有任何问题，上面的updateVelocity函数也能够顺利的找到4个特定观察点的速度。
但是问题来啦，如果并行计算的话，updateVelocity就没法找到这4个特定观察点的速度了。
Starting time loop Time = 0.01 Current Number means: 0.000565 max: 0.0904 smoothSolver: Solving for Ux, Initial residual = 1, Final residual = 2.4007e-06, No Iterations 1 smoothSolver: Solving for Uy, Initial residual = 0.892185, Final residual = 1.24606e-06, No Iterations 1 smoothSolver: Solving for Uz, Initial residual = 0.895991, Final residual = 1.31577e-06, No Iterations 1 GAMG: Solving for p, Initial residual = 1, Final residual = 9.93486e-07, No Iterations 34 time step continuity errors : sum local = 1.12264e-09, global = -1.18173e-10, cumulative = -1.18173e-10 smoothSolver: Solving for epsilon, Initial residual = 1, Final residual = 0.0107252, No Iterations 1 smoothSolver: Solving for k, Initial residual = 1, Final residual = 0.0106817, No Iterations 1 In updateVelocity, number of mesh is 23177 In updateVelocity, number of U is 23177 the velocity on nodes are 4{(0 0 0)} ExecutionTime = 0.57 s ClockTime = 1 s并行和串行用的是同统一套网格，并行是用 scotch方法分成了八个subdomains
值得注意的是：
在串行计算中：In updateVelocity, number of mesh is 184320
在并行计算中：In updateVelocity, number of mesh is 23177
其中23177就是在processor0中的cell数，下面的是在processor0内的U文件FoamFile { version 2.0; format ascii; class volVectorField; location "0.1"; object U; } // * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * // dimensions [0 1 -1 0 0 0 0]; internalField nonuniform List<vector> 23177 ( (0.225112 5.13224e-05 2.5452e-05)也就是说，在并行时这个函数只读入梁processor0里面的速度和网格，其他processor1、2、3... 均没有放进来。
请问各位大神有没有人遇到过类似的情况？
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@huicfd 并行时每个进程完成该进程的任务，这是只输出了0号进程（openfoam好像默认该进程为master prosessor），而其他进程的结果没有输出。要获取其他进程计算得到的数据，需要同时使用open foam中的Pstrean::gatherList函数和Pstream::scatterList函数进行数据同步。
贴一段我的测试代码，希望对你有帮助。if ( Pstream::parRun() ) { List<scalarField> test(Pstream::nProcs()); test[Pstream::myProcNo()].setSize(3, Pstream::myProcNo()); Pstream::gatherList<scalarField>(test); Pstream::scatterList(test); // Pout << test << endl; List<scalar> numcell(Pstream::nProcs()); numcell[Pstream::myProcNo()] = mesh.nCells(); Pstream::gatherList<scalar>(numcell); Pstream::scatterList<scalar>(numcell); // reduce( numcell, sumOp<List<scalar> >() ); reduce( nCells, sumOp<scalar>() ); // Pout << "CPU = " << Pstream::myProcNo() << ", masterProcNo = " // << Pstream::masterNo() << endl; if ( Pstream::master() ) { scalar size = 0.0; for ( label procI=0; procI<Pstream::nProcs(); procI++ ) { size += numcell[procI]; } // Pout << "grid size = " << size << endl; // Pout << "Cell size in each Process is " << numcell // << ", Total cell size = " << nCells << endl; } }
