使用 C 的 OpenCL，run.sh 权限被拒绝

Question

#define _GNU_SOURCE
#include "matmul.h"
#include "util.h"

#define CL_TARGET_OPENCL_VERSION 120
#define CL_USE_DEPRECATED_OPENCL_2_2_APIS
#include <CL/cl.h>
#include <stdio.h>
#include <stdlib.h>

#define CHECK_ERROR(err)                                                       \
  if (err != CL_SUCCESS) {                                                     \
    printf("[%s:%d] OpenCL error %d\n", __FILE__, __LINE__, err);              \
    exit(EXIT_FAILURE);                                                        \
  }

static cl_int err;
static cl_platform_id platform;
static cl_device_id device;
static cl_context context;
static cl_command_queue queue;
static cl_program program;
static cl_kernel kernel;
static cl_mem a_d, b_d, c_d;

void matmul(const float *A, const float *B, float *C, int M, int N, int K) {
  for (int m=0; m<M; m++) {
    for (int n=0; n<N; n++) {
        float acc = 0.0f;
        for (int k=0; k<K; k++) {
            acc += A[k*M + m] * B[n*K + k];
        }
        C[n*M + m] = acc;
    }
  }
}

static void print_platform_info(cl_platform_id platform) {
  size_t sz;
  char *buf;
  CHECK_ERROR(clGetPlatformInfo(platform, CL_PLATFORM_NAME, 0, NULL, &sz));
  buf = (char *)malloc(sz);
  CHECK_ERROR(clGetPlatformInfo(platform, CL_PLATFORM_NAME, sz, buf, NULL));
  printf("Detected OpenCL platform: %s\n", buf);
  free(buf);
}

static void print_device_info(cl_device_id device) {
  size_t sz;
  char *buf;
  CHECK_ERROR(clGetDeviceInfo(device, CL_DEVICE_NAME, 0, NULL, &sz));
  buf = (char *)malloc(sz);
  CHECK_ERROR(clGetDeviceInfo(device, CL_DEVICE_NAME, sz, buf, NULL));
  printf("Detected OpenCL device: %s\n", buf);
  free(buf);
}

static cl_program create_and_build_program_with_source(cl_context context,
                                                       cl_device_id device,
                                                       const char *file_name) {
  FILE *file = fopen(file_name, "rb");
  if (file == NULL) {
    printf("Failed to open %s\n", file_name);
    exit(EXIT_FAILURE);
  }
  fseek(file, 0, SEEK_END);
  size_t source_size = ftell(file);
  rewind(file);
  char *source_code = (char *)malloc(source_size + 1);
  size_t ntotal = 0;
  while (ntotal < source_size) {
    int nread = fread(source_code, sizeof(char), source_size, file);
    ntotal += nread;
  }
  source_code[source_size] = '\0';
  fclose(file);
  cl_program program = clCreateProgramWithSource(
      context, 1, (const char **)&source_code, &source_size, &err);
  CHECK_ERROR(err);
  free(source_code);
  err = clBuildProgram(program, 1, &device, "", NULL, NULL);
  if (err == CL_BUILD_PROGRAM_FAILURE) {
    size_t log_size;
    CHECK_ERROR(clGetProgramBuildInfo(program, device, CL_PROGRAM_BUILD_LOG, 0,
                                      NULL, &log_size));
    char *log = (char *)malloc(log_size + 1);
    CHECK_ERROR(clGetProgramBuildInfo(program, device, CL_PROGRAM_BUILD_LOG,
                                      log_size, log, NULL));
    log[log_size] = 0;
    printf("Compile error:\n%s\n", log);
    free(log);
  }
  CHECK_ERROR(err);
  return program;
}

void matmul_initialize(int M, int N, int K) {
  // Get OpenCL platform
  err = clGetPlatformIDs(1, &platform, NULL);
  CHECK_ERROR(err);
  print_platform_info(platform);

  // Get OpenCL device (only 1)
  err = clGetDeviceIDs(platform, CL_DEVICE_TYPE_GPU, 1, &device, NULL);
  CHECK_ERROR(err);
  print_device_info(device);

  // Create OpenCL context
  context = clCreateContext(NULL, 1, &device, NULL, NULL, &err);
  CHECK_ERROR(err);

  // Create OpenCL command queue
  queue = clCreateCommandQueue(context, device, 0, &err);
  CHECK_ERROR(err);

  // Compile program from "kernel.cl"
  program = create_and_build_program_with_source(context, device, "kernel.cl");

  // Extract kernel from compiled program
  kernel = clCreateKernel(program, "sgemm", &err);
  CHECK_ERROR(err);

  // Create GPU buffers
  a_d = clCreateBuffer(context, CL_MEM_READ_WRITE, M * K * sizeof(float), NULL,
                       &err);
  CHECK_ERROR(err);
  b_d = clCreateBuffer(context, CL_MEM_READ_WRITE, K * N * sizeof(float), NULL,
                       &err);
  CHECK_ERROR(err);
  c_d = clCreateBuffer(context, CL_MEM_READ_WRITE, M * N * sizeof(float), NULL,
                       &err);
  CHECK_ERROR(err);
}

void matmul_finalize() {
  clReleaseMemObject(a_d);
  clReleaseMemObject(b_d);
  clReleaseMemObject(c_d);
  clReleaseCommandQueue(queue);
  clReleaseContext(context);
  clReleaseProgram(program);
  clReleaseKernel(kernel);
}

这个是matmul.c，

#include "util.h"

#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <time.h>

static double start_time[8];

void timer_init() { srand(time(NULL)); }

static double get_time() {
  struct timespec tv;
  clock_gettime(CLOCK_MONOTONIC, &tv);
  return tv.tv_sec + tv.tv_nsec * 1e-9;
}

void timer_start(int i) { start_time[i] = get_time(); }

double timer_stop(int i) { return get_time() - start_time[i]; }

void alloc_mat(float **m, int R, int S) {
  *m = (float *)aligned_alloc(32, sizeof(float) * R * S);
  if (*m == NULL) {
    printf("Failed to allocate memory for mat.\n");
    exit(0);
  }
}

void rand_mat(float *m, int R, int S) {
  int N = R * S;
  for (int j = 0; j < N; j++) {
    m[j] = (float)rand() / RAND_MAX - 0.5;
  }
}

void zero_mat(float *m, int R, int S) {
  int N = R * S;
  memset(m, 0, sizeof(float) * N);
}

void print_mat(float *m, int M, int N) {
  for (int i = 0; i < M; ++i) {
    for (int j = 0; j < N; ++j) {
      printf("%+.3f ", m[i * N + j]);
    }
    printf("\n");
  }
}

void check_mat_mul(float *A, float *B, float *C, int M, int N, int K) {
  printf("Validating...\n");

  float *C_ans;
  alloc_mat(&C_ans, M, N);
  zero_mat(C_ans, M, N);
#pragma omp parallel for
  for (int i = 0; i < M; ++i) {
    for (int k = 0; k < K; ++k) {
      for (int j = 0; j < N; ++j) {
        C_ans[i * N + j] += A[i * K + k] * B[k * N + j];
      }
    }
  }

  bool is_valid = true;
  int cnt = 0, thr = 10;
  float eps = 1e-3;
  for (int i = 0; i < M; ++i) {
    for (int j = 0; j < N; ++j) {
      float c = C[i * N + j];
      float c_ans = C_ans[i * N + j];
      if (fabsf(c - c_ans) > eps &&
          (c_ans == 0 || fabsf((c - c_ans) / c_ans) > eps)) {
        ++cnt;
        if (cnt <= thr)
          printf("C[%d][%d] : correct_value = %f, your_value = %f\n", i, j,
                 c_ans, c);
        if (cnt == thr + 1)
          printf("Too many error, only first %d values are printed.\n", thr);
        is_valid = false;
      }
    }
  }

  if (is_valid) {
    printf("Result: VALID\n");
  } else {
    printf("Result: INVALID\n");
  }
}

这个是utils.c，

#include <getopt.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include "matmul.h"
#include "util.h"

static void print_help(const char *prog_name) {
  printf("Usage: %s [-pvh] [-n num_iterations] M N K\n", prog_name);
  printf("Options:\n");
  printf("     -p : print matrix. (default: off)\n");
  printf("     -v : validate matrix multiplication. (default: off)\n");
  printf("     -h : print this page.\n");
  printf("     -n : number of iterations (default: 1)\n");
  printf("      M : number of rows of matrix A and C. (default: 8)\n");
  printf("      N : number of columns of matrix B and C. (default: 8)\n");
  printf(
      "      K : number of columns of matrix A and rows of B. (default: 8)\n");
}

static bool print_data = false;
static bool validation = false;
static int M = 8;
static int N = 8;
static int K = 8;
static int num_iterations = 1;

static void parse_opt(int argc, char **argv) {
  int c;
  while ((c = getopt(argc, argv, "pvht:n:m:")) != -1) {
    switch (c) {
    case 'p':
      print_data = true;
      break;
    case 'v':
      validation = true;
      break;
    case 'n':
      num_iterations = atoi(optarg);
      break;
    case 'h':
    default:
      print_help(argv[0]);
      exit(0);
    }
  }
  for (int i = optind, j = 0; i < argc; ++i, ++j) {
    switch (j) {
    case 0:
      M = atoi(argv[i]);
      break;
    case 1:
      N = atoi(argv[i]);
      break;
    case 2:
      K = atoi(argv[i]);
      break;
    default:
      break;
    }
  }
  printf("Options:\n");
  printf("  Problem size: M = %d, N = %d, K = %d\n", M, N, K);
  printf("  Number of iterations: %d\n", num_iterations);
  printf("  Print matrix: %s\n", print_data ? "on" : "off");
  printf("  Validation: %s\n", validation ? "on" : "off");
  printf("\n");
}

int main(int argc, char **argv) {
  parse_opt(argc, argv);

  printf("Initializing... ");
  fflush(stdout);
  float *A, *B, *C;

  // Initialize random seed
  timer_init();

  // Allocate matrices
  alloc_mat(&A, M, K);
  alloc_mat(&B, K, N);
  alloc_mat(&C, M, N);

  // Set each element to a random value
  rand_mat(A, M, K);
  rand_mat(B, K, N);

  printf("done!\n");

  // Initialize OpenCL
  printf("Initializing OpenCL...\n");
  fflush(stdout);
  matmul_initialize(M, N, K);

  // Few warmup iterations
  zero_mat(C, M, N);
  for (int i = 0; i < 3; i++) {
    matmul(A, B, C, M, N, K);
  }

  double elapsed_time_sum = 0;
  for (int i = 0; i < num_iterations; ++i) {
    printf("Calculating...(iter=%d) ", i);
    fflush(stdout);
    zero_mat(C, M, N);

    timer_start(0);
    matmul(A, B, C, M, N, K);

    double elapsed_time = timer_stop(0);
    printf("%f sec\n", elapsed_time);
    elapsed_time_sum += elapsed_time;
  }

  if (print_data) {
    printf("MATRIX A:\n");
    print_mat(A, M, K);
    printf("MATRIX B:\n");
    print_mat(B, K, N);
    printf("MATRIX C:\n");
    print_mat(C, M, N);
  }

  // Finalize OpenCL
  matmul_finalize();

  if (validation) {
    check_mat_mul(A, B, C, M, N, K);
  }

  double elapsed_time_avg = elapsed_time_sum / num_iterations;
  printf("Avg. time: %f sec\n", elapsed_time_avg);
  printf("Avg. throughput: %f GFLOPS\n",
         2.0 * M * N * K / elapsed_time_avg / 1e9);

  return 0;
}

这是主要的。c

我通过

运行这段代码

#!/bin/bash

srun --nodes=1 --exclusive --partition=class1 --gres=gpu:4 ./main $@

在 GPU 集群服务器上，但它给了我

bash：./run_performance.sh：权限被拒绝

这个错误。

我认为GPU的访问权限没有问题。

我可以解决这个问题吗？

我不确定但是

#define CL_TARGET_OPENCL_VERSION 120
#define CL_USE_DEPRECATED_OPENCL_2_2_APIS

这个版本的问题似乎是问题。

这是服务器的GPU信息

Number of platforms                               1
  Platform Name                                   NVIDIA CUDA
  Platform Vendor                                 NVIDIA Corporation
  Platform Version                                OpenCL 3.0 CUDA 11.8.88
  Platform Profile                                FULL_PROFILE
  Platform Extensions                             cl_khr_global_int32_base_atomics cl_khr_global_int32_extended_atomics cl_khr_local_int32_base_atomics cl_khr_local_int32_extended_atomics cl_khr_fp64 cl_khr_3d_image_writes cl_khr_byte_addressable_store cl_khr_icd cl_khr_gl_sharing cl_nv_compiler_options cl_nv_device_attribute_query cl_nv_pragma_unroll cl_nv_copy_opts cl_nv_create_buffer cl_khr_int64_base_atomics cl_khr_int64_extended_atomics cl_khr_device_uuid cl_khr_pci_bus_info cl_khr_external_semaphore cl_khr_external_memory cl_khr_external_semaphore_opaque_fd cl_khr_external_memory_opaque_fd
  Platform Host timer resolution                  0ns
  Platform Extensions function suffix             NV

  Platform Name                                   NVIDIA CUDA
Number of devices                                 1
  Device Name                                     NVIDIA TITAN RTX
  Device Vendor                                   NVIDIA Corporation
  Device Vendor ID                                0x10de
  Device Version                                  OpenCL 3.0 CUDA
  Driver Version                                  520.61.05
  Device OpenCL C Version                         OpenCL C 1.2 
  Device Type                                     GPU

更改 matmul.c 的 #define 部分的版本和 API 版本

Answer 1

看起来

run_performance.sh

没有执行文件权限。在Linux中，每个文件都有单独的读/写/执行权限，并且默认情况下执行权限是关闭的，以避免文件的意外执行和对系统的潜在危害。

就您而言，您希望该文件可执行。要解决这个问题，请运行：

chmod +x run_performance.sh

之后你可以执行它：

./run_performance.sh