CUDA双精度和每个线程的寄存器数量

Question

I am having an error while executing the kernel

too many resources requested for launch

I checked online for any hints on error message, which suggest this happens due to usage of more registers than the limit specified by the GPU for each multi-processsor. Device query results as follows:

Device 0: "GeForce GTX 470"
CUDA Driver Version / Runtime Version          5.0 / 5.0
CUDA Capability Major/Minor version number:    2.0
Total amount of global memory:                 1279 MBytes (1341325312 bytes)
(14) Multiprocessors x ( 32) CUDA Cores/MP:    448 CUDA Cores
GPU Clock rate:                                1215 MHz (1.22 GHz)
Memory Clock rate:                             1674 Mhz
Memory Bus Width:                              320-bit
L2 Cache Size:                                 655360 bytes
Total amount of constant memory:               65536 bytes
Total amount of shared memory per block:       49152 bytes
Total number of registers available per block: 32768
Warp size:                                     32
Maximum number of threads per multiprocessor:  1536
Maximum number of threads per block:           1024
Maximum sizes of each dimension of a block:    1024 x 1024 x 64
Maximum sizes of each dimension of a grid:     65535 x 65535 x 65535

Update Robert Crovella remarked that he had no problems in running the code, so I paste here the complete code snippet for execution.

Complete code looks like this:

__global__ void calc_params(double *d_result_array, int total_threads) {

        int thread_id             = threadIdx.x + (blockDim.x * threadIdx.y);
        d_result_array[thread_id] = 1 / d_result_array[thread_id];

 }

  void calculate() {

     double *h_array;
     double *d_array;

     size_t array_size = pow((double)31, 2) * 2 * 10;

     h_array = (double *)malloc(array_size * sizeof(double));
     cudaMalloc((void **)&d_array, array_size * sizeof(double));

     for (int i = 0; i < array_size; i++) {
        h_array[i] = i;
     }

     cudaMemcpy(d_array, h_array, array_size * sizeof(double), cudaMemcpyHostToDevice);

     int BLOCK_SIZE = 1024;
     int NUM_OF_BLOCKS = (array_size / BLOCK_SIZE) + (array_size % BLOCK_SIZE)?1:0;

     calc_params<<<NUM_OF_BLOCKS, BLOCK_SIZE>>>(d_array, array_size);
     cudaDeviceSynchronize();
     checkCudaErrors(cudaGetLastError());

     cudaFree(d_array);
     free(h_array);

}

When I execute this code, I get the error as, too many resources requested for launch

While instead of using the inverse statement in the kernel
(ie d_result_array[thread_id] = 1 / d_result_array[thread_id])
the equate statement works perfectly
(ie d_result_array[thread_id] = d_result_array[thread_id] * 200) .

Why? Is there any possible alternative to that (other than using a smaller block size). If thats the only solution, how shall I know what should be the block size that can work.

Regards,

PS For those who are might wanna know whats cudaCheckErrors is

#define checkCudaErrors(val) check( (val), #val, __FILE__, __LINE__)

template<typename T>
void check(T err, const char* const func, const char* const file, const int line) {
  if (err != cudaSuccess) {
    std::cerr << "CUDA error at: " << file << ":" << line << std::endl;
    std::cerr << cudaGetErrorString(err) << " " << func << std::endl;
    exit(1);
  }
}

Build and OS Information

Build of configuration Debug for project TEST

make all 
Building file: ../test_param.cu
Invoking: NVCC Compiler
nvcc -G -g -O0 -gencode arch=compute_20,code=sm_20 -odir "" -M -o "test_param.d" "../test_param.cu"
nvcc --compile -G -O0 -g -gencode arch=compute_20,code=compute_20 -gencode arch=compute_20,code=sm_20  -x cu -o  "test_param.o" "../test_param.cu"
Finished building: ../test_param.cu

Building target: TEST
Invoking: NVCC Linker
nvcc  -link -o  "TEST"  ./test_param.o   
Finished building target: TEST

Operating System

Ubuntu Lucid (10.04.4) 64bit
Linux paris 2.6.32-46-generic #105-Ubuntu SMP Fri Mar 1 00:04:17 UTC 2013 x86_64 GNU/Linux

Error I receive

CUDA error at: ../test_param.cu:42
too many resources requested for launch cudaGetLastError()

Answer 1

This seems to be an artifact of the compiler. The problem seems to be the register usage, which you can observe by passing the -Xptxas -v option on the nvcc command line. For some reason the -G version of the code uses quite a bit more registers (per thread) than the regular code. You have a few options:

1. Don't use the -G switch. This switch should only be used for debug purposes anyway, as it generates code that may run slower than without the -G switch.
2. If you want to use the -G switch, then reduce the number of threads per block. For the example in this case, I was able to get it to run with 768 threads per block or less.

3. Instruct the compiler to use fewer registers per thread. You can do this with the -maxrregcount switch, such as:

    nvcc -Xptxas -v -arch=sm_20 -G -maxrregcount=20 -o t145 t145.cu 

The objective in this last case is to have the (registers per thread * threads per block) be less than the max registers per block for the GPU in use. A typical CC 2.0 GPU has a maximum of 32768 registers available per block (which you can discover with the deviceQuery sample ).