Question

我的计算机中有一个专用的计算GPU（不用于显示）。它的属性是：

Device 0: "Tesla C2050"
CUDA Driver Version / Runtime Version          6.0 / 6.0
CUDA Capability Major/Minor version number:    2.0
Total amount of global memory:                 2688 MBytes (2818244608 bytes)
(14) Multiprocessors, ( 32) CUDA Cores/MP:     448 CUDA Cores
GPU Clock rate:                                1147 MHz (1.15 GHz)
Memory Clock rate:                             1500 Mhz
Memory Bus Width:                              384-bit
L2 Cache Size:                                 786432 bytes
Maximum Texture Dimension Size (x,y,z)         1D=(65536), 2D=(65536, 65535), 3D=(2048, 2048, 2048)
Maximum Layered 1D Texture Size, (num) layers  1D=(16384), 2048 layers
Maximum Layered 2D Texture Size, (num) layers  2D=(16384, 16384), 2048 layers
Total amount of constant memory:               65536 bytes
Total amount of shared memory per block:       49152 bytes
Total number of registers available per block: 32768
Warp size:                                     32
Maximum number of threads per multiprocessor:  1536
Maximum number of threads per block:           1024
Max dimension size of a thread block (x,y,z): (1024, 1024, 64)
Max dimension size of a grid size    (x,y,z): (65535, 65535, 65535)
Maximum memory pitch:                          2147483647 bytes
Texture alignment:                             512 bytes
Concurrent copy and kernel execution:          Yes with 2 copy engine(s)
Run time limit on kernels:                     No
Integrated GPU sharing Host Memory:            No
Support host page-locked memory mapping:       Yes
Alignment requirement for Surfaces:            Yes
Device has ECC support:                        Enabled
Device supports Unified Addressing (UVA):      Yes

我正在尝试在其上运行以下简单程序（将数组复制到设备）：

#include <cuda.h>
#include <curand_kernel.h>

#define N 252000 

int main( void ) {
    int a[N];
    int *dev_a;
    cudaSetDevice(0);

    cudaMalloc( (void**)&dev_a, N * sizeof(int) );
    for (long i=0; i<N; i++) {
        a[i] = 1;
    }
    cudaMemcpy( dev_a, a, N * sizeof(int), cudaMemcpyHostToDevice ); //**Crashes here**

    cudaFree( dev_a );
    cudaDeviceReset();
    return 0;
}

如果N = 251000该程序有效。但如果N = 252000程序在cudaMemcpy()崩溃。知道为什么会这样吗？

Answer 1

恭喜，您刚刚发现了堆栈大小的限制：

int a[N];

动态分配主机阵列：

int *a = (int *)malloc(N*sizeof(int));

这将从堆中分配。如果你搜索SO，你会发现许多问题，例如解释堆栈与堆分配以及限制的问题。

不明原因的CUDA崩溃

1 个答案: