Why my cuda C code does not become faster with single precision?

Question

Fermi generation GPU's single precision calculation should be 2 times faster than double precision. However, although I rewrite all declaration 'double' to 'float', I got no speed up. Is there any mistake ex. compile option etc..?

GPU:Tesla C2075 OS:win7 pro Compiler:VS2013(nvcc) CUDA:v.7.5 Command line:nvcc test.cu

I wrote test code:

#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

Well, after some investigation, that's because you just perform a multiplication by the constant 1, which gets optimized to "do nothing" in the binary:

If instead you square the array (to prevent this trivial optimization), you get the following assembly:

and the performance gains are restored on the below(simplified) piece of code, in which i changed a few things:

• way larger array (100M)
• using blockDim.x instead of an argument parameter
• use better kernel configuration for my machine (GTX 980)
• allocate input array on heap instead of stack (to allow more than 1M)

here is the code:

#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;
} 

(You'll notice I allocate a array of length 100M to get measurable performance.)