Question

我在Cuda Documentaion中读到，在每个块内部线程都是在一组32个被称为warp的执行中，每个线程指向相同的指令但是可以访问多个数据，我的任务是测试语句的真实性

现在我做的是我发布了一个256线程和一个块的内核，所以8批次必须执行warp。

我将创建一个大小为32的共享变量，将其分配给

sharedVariable [ threadIdx.x % 32 ] = threadIdx.x /32;

然后将该变量分配给256字节长度的全局变量：

outputPointer[ threadIdx.x  ] = sharedVariable [ threadIdx.x % 32 ];

理想情况下，根据假设我应该输出

0,0,0,0,0,0,0,0，直至32 1,1,1,1,1,12 32 .. 2,2,2,2,2，直到32

但我的输出仅为4,4,4,4,4

Cuda Code：

__global__ void addKernel(int *inputPointer, int *outputPointer)
{
    __shared__ int sharedVariable[ 32 ];

    sharedVariable [ threadIdx.x % 32 ] = 0 ;
    sharedVariable [ threadIdx.x % 32 ] = threadIdx.x /32;


    outputPointer[ threadIdx.x  ] = sharedVariable [ threadIdx.x % 32 ];
}

int main () { 
......
 addKernel<<<1, 256>>>(device_inputPointer, device_outputPointer);
......
/**Print output here */
//I am getting 4 ,4,4,4,4,4,4,4,4 as output
}

完整代码：

#include "cuda_runtime.h"
#include "cuda_runtime.h"
#include "device_launch_parameters.h"

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

#include <conio.h>
#define SIZE  256 * sizeof(int)


__global__ void addKernel(int *inputPointer, int *outputPointer)
{
    __shared__ int sharedVariable[ 32 ];


    sharedVariable [ threadIdx.x % 32 ] = 0;

    sharedVariable [ threadIdx.x % 32 ] = threadIdx.x /32;


    outputPointer[ threadIdx.x  ] = sharedVariable [ threadIdx.x % 32 ];
}



int main()
{





    // Copy input vectors from host memory to GPU buffers.
    int *inputPointer = (int * ) malloc (SIZE);
    int *outputPointer= (int * ) malloc (SIZE);


    int *device_inputPointer;
    int *device_outputPointer;

    cudaMalloc((void**)&device_inputPointer, SIZE);
    cudaMalloc((void**)&device_outputPointer, SIZE);

    memset (inputPointer  , 0  ,  SIZE);
    cudaMemcpy(device_inputPointer , inputPointer,  SIZE , cudaMemcpyHostToDevice);

    // Launch a kernel on the GPU with one thread for each element.
    addKernel<<<1, 256>>>(device_inputPointer, device_outputPointer);

    cudaMemcpy(outputPointer, device_outputPointer, SIZE , cudaMemcpyDeviceToHost);

    // cudaDeviceReset must be called before exiting in order for profiling and
    // tracing tools such as Nsight and Visual Profiler to show complete traces.

    for ( int i = 0 ; i < 256  ; i ++ ) {
        printf ( " %d  " , outputPointer[i] );
    }

    cudaDeviceReset();



    getch();
    return 0;
}

#include "device_launch_parameters.h"

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

#include <conio.h>
#define SIZE  256 * sizeof(int)


__global__ void addKernel(int *inputPointer, int *outputPointer)
{
    __shared__ int sharedVariable[ 32 ];


    sharedVariable [ threadIdx.x % 32 ] = 0;

    sharedVariable [ threadIdx.x % 32 ] = threadIdx.x /32;


    outputPointer[ threadIdx.x  ] = sharedVariable [ threadIdx.x % 32 ];
}



int main()
{





    // Copy input vectors from host memory to GPU buffers.
    int *inputPointer = (int * ) malloc (SIZE);
    int *outputPointer= (int * ) malloc (SIZE);


    int *device_inputPointer;
    int *device_outputPointer;

    cudaMalloc((void**)&device_inputPointer, SIZE);
    cudaMalloc((void**)&device_outputPointer, SIZE);

    memset (inputPointer  , 0  ,  SIZE);
    cudaMemcpy(device_inputPointer , inputPointer,  SIZE , cudaMemcpyHostToDevice);

    // Launch a kernel on the GPU with one thread for each element.
    addKernel<<<1, 256>>>(device_inputPointer, device_outputPointer);

    cudaMemcpy(outputPointer, device_outputPointer, SIZE , cudaMemcpyDeviceToHost);

    // cudaDeviceReset must be called before exiting in order for profiling and
    // tracing tools such as Nsight and Visual Profiler to show complete traces.

    for ( int i = 0 ; i < 256  ; i ++ ) {
        printf ( " %d  " , outputPointer[i] );
    }

    cudaDeviceReset();



    getch();
    return 0;
}

我在不同的硬件上测试过它在K20（特斯拉建筑，它工作正常）

Answer 1

代码在以下行中具有未定义的行为：

 sharedVariable [ threadIdx.x % 32 ] = 0;
 sharedVariable [ threadIdx.x % 32 ] = threadIdx.x /32;

多个线程可以具有相同的threadIdx.x % 32值，并且这些线程会尝试同时写入相同的共享内存位置。这将导致这些线程之间的竞争条件。

例如，考虑threadIdx.x 0,32,64,96等的线程。所有这些线程都将尝试访问0的索引sharedVariable，从而导致未定义的行为。所有线程彼此偏移32的情况也是如此（仅在当前示例中）。

Answer 2

我认为你并不完全清楚cuda代码是如何并行执行的。第sharedVariable [ threadIdx.x % 32 ] = 0 ;行完全没用，因为在下一行中，您使用threadIdx.x /32覆盖它。

此外，您对经线计数的假设是错误的。您的threadblock包含256个线程。所以有8个经线（256/32）。

您不能指望您的代码有任何结果，因为没有明确的行为！ threadIdx.x /32的结果将在0到7的范围内，具体取决于范围为0..255的threadIdx.x。因为每32个线程有8个warp，所以将有8个写入sharedVariable [ threadIdx.x % 32 ]并且你没有控制权，warp将首先执行并且最后执行。

在你的情况下，warp 4最后被执行，因此你的结果只有4个。

要获得您期望从内核获得的结果，可以将其更改为：

__global__ void addKernel(int *outputPointer)
{
    outputPointer[ threadIdx.x  ] = threadIdx.x /32;
}

我认为没有机会像你想的那样使用共享内存。

关于CUDA Warps的简单代码

2 个答案: