使用CUBLAS查找最大和最小值 [英] Finding maximum and minimum with CUBLAS

问题描述

我遇到了问题，为什么我的函数在使用CUBLAS的双精度范围内找到最大值和最小值的函数无法正常工作。

I'm having problems grasping why my function that finds maximum and minimum in a range of doubles using CUBLAS doesn't work properly.

代码如下：

void findMaxAndMinGPU(double* values, int* max_idx, int* min_idx, int n)
{
    double* d_values;
    cublasHandle_t handle;
    cublasStatus_t stat;
    safecall( cudaMalloc((void**) &d_values, sizeof(double) * n), "cudaMalloc     (d_values) in findMaxAndMinGPU");
    safecall( cudaMemcpy(d_values, values, sizeof(double) * n, cudaMemcpyHostToDevice), "cudaMemcpy (h_values > d_values) in findMaxAndMinGPU");
    cublasCreate(&handle);

    stat = cublasIdamax(handle, n, d_values, sizeof(double), max_idx);
    if (stat != CUBLAS_STATUS_SUCCESS)
        printf("Max failed\n");

    stat = cublasIdamin(handle, n, d_values, sizeof(double), min_idx);
    if (stat != CUBLAS_STATUS_SUCCESS)
        printf("min failed\n");

    cudaFree(d_values);
    cublasDestroy(handle);
}

其中values是要在其中搜索的值。 max_idx和min_idx是值中找到的数字的索引。
CUBLAS调用的结果似乎相当随机，并输出错误的索引。

Where values is the values to search within. The max_idx and min_idx are the index of the found numbers in values. The results from the CUBLAS-calls seems rather random and output wrong indexes.

任何人对我的问题有一个好的答案？我现在很痛苦：（

Anyone with a golly good answer to my problem? I am a tad sad at the moment :(

推荐答案

cublasIdamax 和 cublasIdamin 调用是错误的。BLAS 1级调用中的 incx 所以我怀疑你想要更像的东西：

One of your arguments to both the cublasIdamax and cublasIdamin calls are wrong. The incx argument in BLAS level 1 calls should always be the stride of the input in words, not bytes. So I suspect that you want something more like:

stat = cublasIdamax(handle, n, d_values, 1, max_idx);
if (stat != CUBLAS_STATUS_SUCCESS)
    printf("Max failed\n");

stat = cublasIdamin(handle, n, d_values, 1, min_idx);
if (stat != CUBLAS_STATUS_SUCCESS)
    printf("min failed\n");

通过使用 sizeof（double），你告诉例程使用8的跨度，我假定你在 d_values 中实际上有1的步幅。

By using sizeof(double) you are telling the routines to use a stride of 8, which will have the calls overrun the allocated storage of the input array and into uninitialised memory. I presume you actually have a stride of 1 in d_values.

编辑：这是一个完整的可运行示例，可以正常工作。注意我把代码切换到单精度，因为我目前不能访问双精度的硬件：

Here is a complete runnable example which works correctly. Note I switched the code to single precision because I don't presently have access to double precision capable hardware:

#include <cuda_runtime.h>
#include <cublas_v2.h>
#include <cstdio>
#include <cstdlib>
#include <sys/time.h>


typedef float Real;

void findMaxAndMinGPU(Real* values, int* max_idx, int* min_idx, int n)
{
    Real* d_values;
    cublasHandle_t handle;
    cublasStatus_t stat;
    cudaMalloc((void**) &d_values, sizeof(Real) * n);
    cudaMemcpy(d_values, values, sizeof(Real) * n, cudaMemcpyHostToDevice);
    cublasCreate(&handle);

    stat = cublasIsamax(handle, n, d_values, 1, max_idx);
    if (stat != CUBLAS_STATUS_SUCCESS)
        printf("Max failed\n");

    stat = cublasIsamin(handle, n, d_values, 1, min_idx);
    if (stat != CUBLAS_STATUS_SUCCESS)
        printf("min failed\n");

    cudaFree(d_values);
    cublasDestroy(handle);
}

int main(void)
{
    const int vmax=1000, nvals=10000;

    float vals[nvals];
    srand ( time(NULL) );
    for(int j=0; j<nvals; j++) {
       vals[j] = float(rand() % vmax);
    }

    int minIdx, maxIdx;
    findMaxAndMinGPU(vals, &maxIdx, &minIdx, nvals);

    int cmin = 0, cmax=0;
    for(int i=1; i<nvals; i++) {
        cmin = (vals[i] < vals[cmin]) ? i : cmin;
        cmax = (vals[i] > vals[cmax]) ? i : cmax;
    }

    fprintf(stdout, "%d %d %d %d\n", minIdx, cmin, maxIdx, cmax);

    return 0;
}

在编译和运行时提供：

$ g++ -I/usr/local/cuda/include -L/usr/local/cuda/lib cublastest.cc -lcudart -lcublas
$ ./a.out
273 272 85 84

请注意，CUBLAS遵循FORTRAN惯例，使用1个索引，而不是零索引，这就是为什么CUBLAS和CPU版本之间的差异为1的原因。

note that CUBLAS follows the FORTRAN convention and uses 1 indexing, rather than zero indexing, which is why there is a difference of 1 between the CUBLAS and CPU versions.

这篇关于使用CUBLAS查找最大和最小值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！