c - 为什么我的 cuda C 代码在单精度下没有变得更快？

Question

Fermi 一代 GPU 的单精度计算应该比 double 快 2 倍。然而，尽管我将所有声明“double”重写为“float”，但我并没有加速。有什么错误吗？编译选项等..?

显卡:特斯拉C2075操作系统:win7专业版编译器:VS2013(nvcc)CUDA:v.7.5命令行:nvcc test.cu

我写了测试代码:

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<time.h>
#include<conio.h>

#include<cuda_runtime.h>
#include<cuda_profiler_api.h> 
#include<device_functions.h>
#include<device_launch_parameters.h>

#define DOUBLE 1

#define MAXI 10

__global__ void Kernel_double(double*a,int nthreadx)
{
    double b=1.e0;
    int i;
    i = blockIdx.x * nthreadx + threadIdx.x + 0;
    a[i] *= b;
}
__global__ void Kernel_float(float*a,int nthreadx)
{
    float b=1.0F;
    int i;
    i = blockIdx.x * nthreadx + threadIdx.x + 0;
    a[i] *= b;
}

int main()
{
#if DOUBLE
    double a[10];
    for(int i=0;i<MAXI;++i){
        a[i]=1.e0;
    }
    double*d_a;
    cudaMalloc((void**)&d_a, sizeof(double)*(MAXI));
    cudaMemcpy(d_a, a, sizeof(double)*(MAXI), cudaMemcpyHostToDevice);
#else
    float a[10];
    for(int i=0;i<MAXI;++i){
        a[i]=1.0F;
    }
    float*d_a;
    cudaMalloc((void**)&d_a, sizeof(float)*(MAXI));
    cudaMemcpy(d_a, a, sizeof(float)*(MAXI), cudaMemcpyHostToDevice);
#endif

    dim3 grid(2, 2, 1);
    dim3 block(2, 2, 1);

    clock_t start_clock, end_clock;
    double sec_clock;

    printf("[%d] start\n", __LINE__);
    start_clock = clock();

    for (int i = 1; i <= 100000; ++i){
#if DOUBLE
        Kernel_double << < grid, block >> > (d_a, 2);
        cudaMemcpy(a, d_a, sizeof(double)*(MAXI), cudaMemcpyDeviceToHost);
#else
        Kernel_float << < grid, block >> > (d_a, 2);
        cudaMemcpy(a, d_a, sizeof(float)*(MAXI), cudaMemcpyDeviceToHost);
#endif
    }

    end_clock = clock();
    sec_clock = (end_clock - start_clock) / (double)CLOCKS_PER_SEC;
    printf("[%d] %f[s]\n", __LINE__, sec_clock);
    printf("[%d] end\n", __LINE__);

    return 0;
}

Answer 1

好吧，经过一些调查，那是因为你只是乘以常数 1，它在二进制文件中被优化为“什么都不做”:

如果改为对数组进行平方(以防止这种微不足道的优化)，则会得到以下程序集:

下面(简化的)代码恢复了性能提升，我在其中做了一些更改:

• 更大的数组 (100M)
• 使用 blockDim.x 而不是实参参数
• 为我的机器使用更好的内核配置 (GTX 980)
• 在堆上而不是堆栈上分配输入数组(允许超过 1M)

代码如下:

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<time.h>
#include<conio.h>

#include<cuda_runtime.h>
#include<cuda_profiler_api.h> 
#include<device_functions.h>
#include<device_launch_parameters.h>

#define DOUBLE float

#define ITER 10
#define MAXI 100000000

__global__ void kernel(DOUBLE*a)
{
    for(int i = blockIdx.x * blockDim.x + threadIdx.x ; i < MAXI; i += blockDim.x * gridDim.x) 
    {
        a[i] *= a[i];
    }
}

int main()
{
    DOUBLE* a = (DOUBLE*) malloc(MAXI*sizeof(DOUBLE));
    for(int i=0;i<MAXI;++i)
    {
        a[i]=(DOUBLE)1.0;
    }
    DOUBLE* d_a;
    cudaMalloc((void**)&d_a, sizeof(DOUBLE)*(MAXI));
    cudaMemcpy(d_a, a, sizeof(DOUBLE)*(MAXI), cudaMemcpyHostToDevice);

    clock_t start_clock, end_clock;
    double sec_clock;

    printf("[%d] start\n", __LINE__);
    start_clock = clock();

    for (int i = 1; i <= ITER; ++i){
        kernel <<< 32, 256>>> (d_a);
    }
    cudaDeviceSynchronize();

    end_clock = clock();
    cudaMemcpy(a, d_a, sizeof(DOUBLE)*(MAXI), cudaMemcpyDeviceToHost);
    sec_clock = (end_clock - start_clock) / (double)CLOCKS_PER_SEC;
    printf("[%d] %f/%d[s]\n", __LINE__, sec_clock, CLOCKS_PER_SEC);
    printf("[%d] end\n", __LINE__);

    return 0;
} 

(您会注意到我分配了一个长度为 100M 的数组以获得可衡量的性能。)