CUDA:cuda设备同步返回错误代码 30

Question

下面的代码是关于Sha-1算法的。

内核中有一个很大的循环（HashRounds=0x40000，262144次）。现在的情况是：

1.在调试模式下运行时，它将报告错误30.但是如果我减少周期时间，例如50000次，则不会报告任何错误。

2.在释放模式下运行时，这是正常的，但是当我增加线程规模时，例如块= 48，线程= 192，它会报告相同的问题。

环境：GTX560Ti+Win8+Visual Studio 2012+Cuda5.5

恳求你的帮助！！

#include <stdio.h>
#include <string.h>
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
typedef unsigned int    uint32;
typedef unsigned char   byte;
typedef struct {
    uint32 state[5];
    uint32 count[2];
    byte buffer[64];
    byte workspace[64]; // Temporary buffer.
} hash_context;
#define rol(value, bits) (((value) << (bits)) | ((value) >> (32 - (bits))))
/* blk0() and blk() perform the initial expand. */
/* I got the idea of expanding during the round function from SSLeay */
#define blk0(i) (block->l[i] = (rol(block->l[i],24)&0xFF00FF00) 
    |(rol(block->l[i],8)&0x00FF00FF))
#define blk(i) (block->l[i&15] = rol(block->l[(i+13)&15]^block->l[(i+8)&15] 
    ^block->l[(i+2)&15]^block->l[i&15],1))
/* (R0+R1), R2, R3, R4 are the different operations used in SHA1 */
#define R0(v,w,x,y,z,i) {z+=((w&(x^y))^y)+blk0(i)+0x5A827999+rol(v,5);w=rol(w,30);}
#define R1(v,w,x,y,z,i) {z+=((w&(x^y))^y)+blk(i)+0x5A827999+rol(v,5);w=rol(w,30);}
#define R2(v,w,x,y,z,i) {z+=(w^x^y)+blk(i)+0x6ED9EBA1+rol(v,5);w=rol(w,30);}
#define R3(v,w,x,y,z,i) {z+=(((w|x)&y)|(w&x))+blk(i)+0x8F1BBCDC+rol(v,5);w=rol(w,30);}
#define R4(v,w,x,y,z,i) {z+=(w^x^y)+blk(i)+0xCA62C1D6+rol(v,5);w=rol(w,30);}
cudaError_t addWithCuda();
__global__ void cryptKernel();
__device__ void hash_initial(hash_context* context);
__device__ void hash_process( hash_context * context, byte * data, size_t len);
__device__ void SHA1Transform(uint32 state[5], byte workspace[64], byte buffer[64]);
int main()
{
    cudaError_t cudaStatus = addWithCuda();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "addWithCuda failed!");
        return 1;
    }
    // cudaDeviceReset must be called before exiting in order for profiling and
    // tracing tools such as Nsight and Visual Profiler to show complete traces.
    cudaStatus = cudaDeviceReset();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaDeviceReset failed!");
        return 1;
    }
    printf("over");
    getchar();
    return 0;
}
// Helper function for using CUDA to add vectors in parallel.
cudaError_t addWithCuda()
{
    cudaError_t cudaStatus;
    // Choose which GPU to run on, change this on a multi-GPU system.
    cudaStatus = cudaSetDevice(0);
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaSetDevice failed!  Do you have a CUDA-capable GPU installed?");
        goto Error;
    }
    cryptKernel<<<1,1>>>();
    // Check for any errors launching the kernel
    cudaStatus = cudaGetLastError();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "addKernel launch failed: %sn", cudaGetErrorString(cudaStatus));
        goto Error;
    }
    // cudaDeviceSynchronize waits for the kernel to finish, and returns
    // any errors encountered during the launch.
    cudaStatus = cudaDeviceSynchronize();
    if (cudaStatus != cudaSuccess) {
        fprintf(stderr, "cudaDeviceSynchronize returned error code %d after launching addKernel!n", cudaStatus);
        goto Error;
    }
Error:
    return cudaStatus;
}
__global__ void cryptKernel()
{
    int i = blockIdx.x*blockDim.x+threadIdx.x;
    byte RawPsw[24] = {''};
    hash_context c;
    hash_initial(&c);
    const int HashRounds=0x40000;
    for (int I=0;I<HashRounds;I++)
    {
        hash_process( &c, RawPsw, 24);
    }
}
__device__ void SHA1Transform(uint32 state[5], byte workspace[64], byte buffer[64])
{
    uint32 a, b, c, d, e;
    typedef union {
        byte c[64];
        uint32 l[16];
    } CHAR64LONG16;
    CHAR64LONG16* block;
    block = (CHAR64LONG16*)buffer;
    /* Copy context->state[] to working vars */
    a = state[0];
    b = state[1];
    c = state[2];
    d = state[3];
    e = state[4];
    /* 4 rounds of 20 operations each. Loop unrolled. */
    R0(a,b,c,d,e, 0); R0(e,a,b,c,d, 1); R0(d,e,a,b,c, 2); R0(c,d,e,a,b, 3);
    R0(b,c,d,e,a, 4); R0(a,b,c,d,e, 5); R0(e,a,b,c,d, 6); R0(d,e,a,b,c, 7);
    R0(c,d,e,a,b, 8); R0(b,c,d,e,a, 9); R0(a,b,c,d,e,10); R0(e,a,b,c,d,11);
    R0(d,e,a,b,c,12); R0(c,d,e,a,b,13); R0(b,c,d,e,a,14); R0(a,b,c,d,e,15);
    R1(e,a,b,c,d,16); R1(d,e,a,b,c,17); R1(c,d,e,a,b,18); R1(b,c,d,e,a,19);
    R2(a,b,c,d,e,20); R2(e,a,b,c,d,21); R2(d,e,a,b,c,22); R2(c,d,e,a,b,23);
    R2(b,c,d,e,a,24); R2(a,b,c,d,e,25); R2(e,a,b,c,d,26); R2(d,e,a,b,c,27);
    R2(c,d,e,a,b,28); R2(b,c,d,e,a,29); R2(a,b,c,d,e,30); R2(e,a,b,c,d,31);
    R2(d,e,a,b,c,32); R2(c,d,e,a,b,33); R2(b,c,d,e,a,34); R2(a,b,c,d,e,35);
    R2(e,a,b,c,d,36); R2(d,e,a,b,c,37); R2(c,d,e,a,b,38); R2(b,c,d,e,a,39);
    R3(a,b,c,d,e,40); R3(e,a,b,c,d,41); R3(d,e,a,b,c,42); R3(c,d,e,a,b,43);
    R3(b,c,d,e,a,44); R3(a,b,c,d,e,45); R3(e,a,b,c,d,46); R3(d,e,a,b,c,47);
    R3(c,d,e,a,b,48); R3(b,c,d,e,a,49); R3(a,b,c,d,e,50); R3(e,a,b,c,d,51);
    R3(d,e,a,b,c,52); R3(c,d,e,a,b,53); R3(b,c,d,e,a,54); R3(a,b,c,d,e,55);
    R3(e,a,b,c,d,56); R3(d,e,a,b,c,57); R3(c,d,e,a,b,58); R3(b,c,d,e,a,59);
    R4(a,b,c,d,e,60); R4(e,a,b,c,d,61); R4(d,e,a,b,c,62); R4(c,d,e,a,b,63);
    R4(b,c,d,e,a,64); R4(a,b,c,d,e,65); R4(e,a,b,c,d,66); R4(d,e,a,b,c,67);
    R4(c,d,e,a,b,68); R4(b,c,d,e,a,69); R4(a,b,c,d,e,70); R4(e,a,b,c,d,71);
    R4(d,e,a,b,c,72); R4(c,d,e,a,b,73); R4(b,c,d,e,a,74); R4(a,b,c,d,e,75);
    R4(e,a,b,c,d,76); R4(d,e,a,b,c,77); R4(c,d,e,a,b,78); R4(b,c,d,e,a,79);
    /* Add the working vars back into context.state[] */
    state[0] += a;
    state[1] += b;
    state[2] += c;
    state[3] += d;
    state[4] += e;
}
__device__ void hash_initial(hash_context* context)
{
    /* SHA1 initialization constants */
    context->state[0] = 0x67452301;
    context->state[1] = 0xEFCDAB89;
    context->state[2] = 0x98BADCFE;
    context->state[3] = 0x10325476;
    context->state[4] = 0xC3D2E1F0;
    context->count[0] = context->count[1] = 0;
}
/* Run your data through this. */
__device__ void hash_process( hash_context * context, byte * data, size_t len)
{
    unsigned int i, j;
    uint32 blen = ((uint32)len)<<3;
    j = (context->count[0] >> 3) & 63;
    if ((context->count[0] += blen) < blen ) context->count[1]++;
    context->count[1] += (uint32)(len >> 29);
    if ((j + len) > 63) {
        memcpy(&context->buffer[j], data, (i = 64-j));
        SHA1Transform(context->state, context->workspace, context->buffer);
        for ( ; i + 63 < len; i += 64) {
            SHA1Transform(context->state, context->workspace, &data[i]);
        }
        j = 0;
    }
    else i = 0;
    if (len > i)
      memcpy(&context->buffer[j], &data[i], len - i);
}

Answer 1

当我增加循环时间或线程规模时，它会报告错误代码 30.It 这意味着内核执行时间太长。因此，它击中了Windows TDR机制。关闭TDR将解决此问题。