Question

I am using 63 registers/thread ,so (32768 is maximum) i can use about 520 threads.I am using now 512 threads in this example.

(The parallelism is in the function “computeEvec” inside global computeEHfields function function.)
The problems are:

1) The mem check error below.

2) When i use numPointsRp>2000 it show me “out of memory” ,but (if i am not doing wrong) i compute the global memory and it’s ok.

——————————-UPDATED—————————

i run the program with cuda-memcheck and it gives me (only when numPointsRs>numPointsRp):

========= Invalid global read of size 4

========= at 0x00000428 in computeEHfields

========= by thread (2,0,0) in block (0,0,0)

========= Address 0x4001076e0 is out of bounds

=========
========= Invalid global read of size 4

========= at 0x00000428 in computeEHfields

========= by thread (1,0,0) in block (0,0,0)

========= Address 0x4001076e0 is out of bounds

=========
========= Invalid global read of size 4

========= at 0x00000428 in computeEHfields

========= by thread (0,0,0) in block (0,0,0)

========= Address 0x4001076e0 is out of bounds

ERROR SUMMARY: 160 errors

———–EDIT—————————-

Also , some times (if i use only threads and not blocks (i haven’t test it for blocks) ) if for example i have numPointsRs=1000 and numPointsRp=100 and then change the numPointsRp=200 and then again change the numPointsRp=100 i am not taking the first results!

import pycuda.gpuarray as gpuarray
import pycuda.autoinit
from pycuda.compiler import SourceModule
import numpy as np
import cmath
import pycuda.driver as drv


Rs=np.zeros((numPointsRs,3)).astype(np.float32)
for k in range (numPointsRs): 
    Rs[k]=[0,k,0]

Rp=np.zeros((numPointsRp,3)).astype(np.float32)
for k in range (numPointsRp): 
    Rp[k]=[1+k,0,0]


#---- Initialization and passing(allocate memory and transfer data) to GPU -------------------------
Rs_gpu=gpuarray.to_gpu(Rs)
Rp_gpu=gpuarray.to_gpu(Rp)


J_gpu=gpuarray.to_gpu(np.ones((numPointsRs,3)).astype(np.complex64))
M_gpu=gpuarray.to_gpu(np.ones((numPointsRs,3)).astype(np.complex64))

Evec_gpu=gpuarray.to_gpu(np.zeros((numPointsRp,3)).astype(np.complex64))
Hvec_gpu=gpuarray.to_gpu(np.zeros((numPointsRp,3)).astype(np.complex64))
All_gpu=gpuarray.to_gpu(np.ones(numPointsRp).astype(np.complex64))


mod =SourceModule("""
#include <pycuda-complex.hpp>
#include <cmath>
#include <vector>
#define RowRsSize %(numrs)d
#define RowRpSize %(numrp)d


typedef  pycuda::complex<float> cmplx;
extern "C"{


    __device__ void computeEvec(float Rs_mat[][3], int numPointsRs,   
         cmplx J[][3],
         cmplx M[][3],
         float *Rp,
         cmplx kp, 
         cmplx eta,
         cmplx *Evec,
         cmplx *Hvec, cmplx *All)

{

    while (c<numPointsRs){
        ...         
                c++;

                }     
        }


__global__  void computeEHfields(float *Rs_mat_, int numPointsRs,   
        float *Rp_mat_, int numPointsRp,    
    cmplx *J_,
    cmplx *M_,
    cmplx  kp, 
    cmplx  eta,
    cmplx E[][3],
    cmplx H[][3], cmplx *All )
    {
        float Rs_mat[RowRsSize][3];
        float Rp_mat[RowRpSize][3];

        cmplx J[RowRsSize][3];
        cmplx M[RowRsSize][3];


    int k=threadIdx.x+blockIdx.x*blockDim.x;

      while (k<numPointsRp)  
     {

        computeEvec( Rs_mat, numPointsRs,  J, M, Rp_mat[k], kp, eta, E[k], H[k], All );
        k+=blockDim.x*gridDim.x;


    }

}
}

"""% { "numrs":numPointsRs, "numrp":numPointsRp},no_extern_c=1)


func = mod.get_function("computeEHfields")


func(Rs_gpu,np.int32(numPointsRs),Rp_gpu,np.int32(numPointsRp),J_gpu, M_gpu, np.complex64(kp), np.complex64(eta),Evec_gpu,Hvec_gpu, All_gpu, block=(128,1,1),grid=(200,1))

print(" \n")


#----- get data back from GPU-----
Rs=Rs_gpu.get()
Rp=Rp_gpu.get()
J=J_gpu.get()
M=M_gpu.get()
Evec=Evec_gpu.get()
Hvec=Hvec_gpu.get()
All=All_gpu.get()

——————–GPU MODEL————————————————

Device 0: "GeForce GTX 560"
  CUDA Driver Version / Runtime Version          4.20 / 4.10
  CUDA Capability Major/Minor version number:    2.1
  Total amount of global memory:                 1024 MBytes (1073283072 bytes)
  ( 0) Multiprocessors x (48) CUDA Cores/MP:     0 CUDA Cores   //CUDA Cores    336 => 7 MP and 48 Cores/MP

Answer 1

When i use numPointsRp>2000 it show me “out of memory”

Now we have some real code to work with, let’s compile it and see what happens. Using RowRsSize=2000 and RowRpSize=200 and compiling with the CUDA 4.2 toolchain, I get:

nvcc -arch=sm_21 -Xcompiler="-D RowRsSize=2000 -D RowRpSize=200" -Xptxas="-v" -c -I./ kivekset.cu 
ptxas info    : Compiling entry function '_Z15computeEHfieldsPfiS_iPN6pycuda7complexIfEES3_S2_S2_PA3_S2_S5_S3_' for 'sm_21'
ptxas info    : Function properties for _Z15computeEHfieldsPfiS_iPN6pycuda7complexIfEES3_S2_S2_PA3_S2_S5_S3_
    122432 bytes stack frame, 0 bytes spill stores, 0 bytes spill loads
ptxas info    : Used 57 registers, 84 bytes cmem[0], 168 bytes cmem[2], 76 bytes cmem[16]

The key numbers are 57 registers and 122432 bytes stack frame per thread. The occupancy calculator suggests that a block of 512 threads will have a maximum of 1 block per SM, and your GPU has 7 SM. This gives a total of 122432 * 512 * 7 = 438796288 bytes of stack frame (local memory) to run your kernel, before you have allocated a single of byte of memory for input and output using pyCUDA. On a GPU with 1Gb of memory, it isn’t hard to imagine running out of memory. Your kernel has a enormous local memory footprint. Start thinking about ways to reduce it.

---

As I indicated in comments, it is absolutely unclear why every thread needs a complete copy of the input data in this kernel code. It results in a gigantic local memory footprint and there seems to be absolutely no reason why the code should be written in this way. You could, I suspect, modify the kernel to something like this:

typedef  pycuda::complex<float> cmplx;
typedef float fp3[3];
typedef cmplx cp3[3];

__global__  
void computeEHfields2(
        float *Rs_mat_, int numPointsRs,
        float *Rp_mat_, int numPointsRp,
        cmplx *J_,
        cmplx *M_,
        cmplx  kp, 
        cmplx  eta,
        cmplx E[][3],
        cmplx H[][3], 
        cmplx *All )
{

    fp3 * Rs_mat = (fp3 *)Rs_mat_;
    cp3 * J = (cp3 *)J_;
    cp3 * M = (cp3 *)M_;

    int k=threadIdx.x+blockIdx.x*blockDim.x;
    while (k<numPointsRp)  
    {
        fp3 * Rp_mat = (fp3 *)(Rp_mat_+k);
        computeEvec2( Rs_mat, numPointsRs, J, M, *Rp_mat, kp, eta, E[k], H[k], All );
        k+=blockDim.x*gridDim.x;
    }
}

and the main __device__ function it calls to something like this:

__device__ void computeEvec2(
        fp3 Rs_mat[], int numPointsRs,   
        cp3 J[],
        cp3 M[],
        fp3   Rp,
        cmplx kp, 
        cmplx eta,
        cmplx *Evec,
        cmplx *Hvec, 
        cmplx *All)
{
 ....
}

and eliminate every byte of thread local memory without changing the functionality of the computational code at all.