2016-04-22 93 views
1

我一直在玩一個簡單的CUDA程序,它只消除全局內存。下面是設備代碼,以及主機代碼:CUDA地址超出界限

#include <stdio.h> 

__global__ void kernel(float *data, int width) { 
    int x = blockDim.x * blockIdx.x + threadIdx.x; 
    int y = blockDim.y * blockIdx.y + threadIdx.y; 

    if (x > (width-1)) { 
     printf("x = %d\n", x); 
     printf("blockDim.x = %d\n", blockDim.x); 
     printf("blockIdx.x = %d\n", blockIdx.x); 
     printf("threadIdx.x = %d\n", threadIdx.x); 
    } 

    if (y > (width-1)) { 
     printf("y = %d\n", y); 
     printf("blockDim.y = %d\n", blockDim.y); 
     printf("blockIdx.y = %d\n", blockIdx.y); 
     printf("threadIdx.y = %d\n", threadIdx.y); 
    } 

    data[y * width + x] = 0.0; 
} 

int main(void) { 
    const int MATRIX_SIZE = 256; 
    float *data, *dataGPU; 
    int sizeOfMem; 
    int x = MATRIX_SIZE; 
    int y = MATRIX_SIZE; 

    cudaDeviceReset(); 
    cudaDeviceSynchronize(); 

    sizeOfMem = sizeof(float) * x * y; 

    data = (float *)malloc(sizeOfMem); 
    cudaMalloc((void **)&dataGPU, sizeOfMem); 

    cudaMemcpy(dataGPU, data, sizeOfMem, cudaMemcpyHostToDevice); 

    //int threads = 256; 
    //int blocks = ((x * y) + threads - 1)/threads; 

    dim3 threads(16, 16); 
    dim3 blocks(x/16, y/16); 

    kernel<<<blocks, threads>>>(dataGPU, MATRIX_SIZE); 
    cudaThreadSynchronize(); 

    cudaMemcpy(data, dataGPU, sizeOfMem, cudaMemcpyDeviceToHost); 

    cudaFree(dataGPU); 

    free(data); 

    return 0; 
} 

我繼續接收地址越界錯誤信息與CUDA的MEMCHECK運行我的代碼的時候。但是,只有矩陣我創建的維度是128或更大。如果我的維度小於128,那麼錯誤頻率就會降低(我幾乎從不會收到錯誤)。您可能會注意到我在我的內核函數中包含了打印語句。只有當我收到錯誤消息時纔會打印這些語句,因爲x和y不應該大於width-1,或者在這種情況下爲255.如果我正確地完成了我的數學運算(我相信自己有這個數字),這種說法是正確的。下面是我從CUDA-MEMCHECK收到一條錯誤消息:

========= CUDA-MEMCHECK 
    ========= Invalid __global__ write of size 4 
    =========  at 0x00000298 in kernel(float*, int) 
    =========  by thread (3,10,0) in block (15,1,0) 
    =========  Address 0x2300da6bcc is out of bounds 
    =========  Saved host backtrace up to driver entry point at kernel launch time 
    =========  Host Frame:/usr/lib64/nvidia/libcuda.so.1 (cuLaunchKernel + 0x2c5) [0x472225] 
    =========  Host Frame:./test_reg_memory [0x16c41] 
    =========  Host Frame:./test_reg_memory [0x31453] 
    =========  Host Frame:./test_reg_memory [0x276d] 
    =========  Host Frame:./test_reg_memory [0x24f0] 
    =========  Host Frame:/lib64/libc.so.6 (__libc_start_main + 0xf5) [0x21b15] 
    =========  Host Frame:./test_reg_memory [0x25cd] 
    ========= 
    y = 2074 
    blockDim.y = 16 
    blockIdx.y = 1 
    threadIdx.y = 10 

這個輸出是沒有意義的我,因爲如果我做數學題,

y = blockDim.y * blockIdx.y + threadIdx.y = 16 * 1 + 10 = 26 (not 2074) 

我花了一些時間在CUDA編程論壇,似乎沒有任何幫助。我讀過一個線程,表明我可能會損壞寄存器內存。然而,開始線程的這個問題與不同的GPU有關。線程有點不相關,但我總是包含鏈接。

https://devtalk.nvidia.com/default/topic/498784/memory-corruption-on-a-fermi-class-gpu-error-only-on-fermis-program-works-on-non-fermis-/?offset=6

下面我已經包括NVCC版本。

nvcc: NVIDIA (R) Cuda compiler driver 
Copyright (c) 2005-2015 NVIDIA Corporation 
Built on Tue_Aug_11_14:27:32_CDT_2015 
Cuda compilation tools, release 7.5, V7.5.17 

此外,這裏是我正在使用的GPU。

Device 0: "GeForce GT 640" 
CUDA Driver Version/Runtime Version 8.0/7.5 
CUDA Capability Major/Minor version number: 3.0 

任何有CUDA經驗的人都會指出我可能做錯了什麼嗎?

+1

您發佈的代碼對我來說運行正常,並且不會在cuda-memcheck中產生任何錯誤。如果您從SO問題中複製粘貼,編譯並運行它,您真的確定發佈的代碼會提供cuda-memcheck錯誤嗎? – talonmies

+0

cudaMalloc是否成功? –

+0

@RegisPortalez:如果cudaMalloc失敗,cuda-memcheck會報告錯誤。發佈的輸出不包含此類錯誤。 – talonmies

回答

0

這個問題似乎被限制在一個特定的系統中,並且由某種硬件問題引起。代碼本身很好,並且更改爲不同的系統證實其正常工作。

[此答案已從評論中彙總,並作爲社區wiki條目添加,以將他的問題從CUDA標記的未答覆隊列中移除]。