cuda(1) 最大并发量

Created on 2013-8-5
URL : http://blog.sina.com.cn/s/blog_a502f1a30101mi6t.html
@author: zhxfl
转载请说明出处

cudaDeviceProp  prop;

    int count;
    ( cudaGetDeviceCount( &count ) );
    for (int i=0; i< count; i++) {
        ( cudaGetDeviceProperties( &prop, i ) );
        printf( "   --- General Information for device %d ---
", i );
        printf( "Name:  %s
", prop.name );
        printf( "Compute capability:  %d.%d
", prop.major, prop.minor );
        printf( "Clock rate:  %d
", prop.clockRate );
        printf( "Device copy overlap:  " );
        if (prop.deviceOverlap)
            printf( "Enabled
" );
        else
            printf( "Disabled
");
        printf( "Kernel execution timeout :  " );
        if (prop.kernelExecTimeoutEnabled)
            printf( "Enabled
" );
        else
            printf( "Disabled
" );

        printf( "   --- Memory Information for device %d ---
", i );
        printf( "Total global mem:  %ld
", prop.totalGlobalMem );
        printf( "Total constant Mem:  %ld
", prop.totalConstMem );
        printf( "Max mem pitch:  %ld
", prop.memPitch );
        printf( "Texture Alignment:  %ld
", prop.textureAlignment );

        printf( "   --- MP Information for device %d ---
", i );
        printf( "Multiprocessor count:  %d
",
            prop.multiProcessorCount );
        printf( "Shared mem per mp:  %ld
", prop.sharedMemPerBlock );
        printf( "Registers per mp:  %d
", prop.regsPerBlock );
        printf( "Threads in warp:  %d
", prop.warpSize );
        printf( "Max threads per block:  %d
",
            prop.maxThreadsPerBlock );
        printf( "Max thread dimensions:  (%d, %d, %d)
",
            prop.maxThreadsDim[0], prop.maxThreadsDim[1],
            prop.maxThreadsDim[2] );
        printf( "Max grid dimensions:  (%d, %d, %d)
",
            prop.maxGridSize[0], prop.maxGridSize[1],
            prop.maxGridSize[2] );
        printf( "
" );
    }

    int n1 = rand() % base + base;
    int m1 = rand() % base + base;
    int n2 = m1;
    int m2 = rand() % base + base;
    int *g1 = new int[n1 * m1];
    int *g2 = new int[n2 * m2];
    printf("matrix A[%3d %3d]
", n1, m1);
    for(int i = 0; i < n1 * m1;i++)
    {
        g1[i] = rand() % large;
        //printf("%5d ", g1[i]);
        //if((i + 1) % m1 == 0)printf("
");
    }
    printf("matrix B[%3d %3d]
", n2, m2);
    for(int i = 0; i < n2 * m2;i++)
    {
        g2[i] = rand() % large;
        //printf("%5d ", g2[i]);
        //if((i + 1) % m2 == 0)printf("
");
    }
    int *g;


    g = matrixMultiply(g1,n1,m1,g2,n2,m2);

    printf("matrix C[%3d %3d]
", n1, m2);
    for(int i = 0; i< n1*m2;i++)
    {
        //printf("%5d ", g[i]);
        //if((i + 1) % m2 == 0) printf("
");
    }

上面是cuda example的代码，其中maxThreadsPerBlock是指每个块上的最大线程数，maxGridSize是最大的blocks数。理论上讲最大的并发量是maxThreadsPerBlock  * maxGridSize。下面是在我本机上的运行结果，可以看到65535 * 1024是我想要的答案。

下面来测试一下

1）崩溃：function <<<65535,1024>>()

2）正常：function <<<65536,1>>()

3）崩溃：function <<<65536,1>>()

4）正常：function <<<1,1024>>()

5）崩溃：function <<<1,1025>>()

其中第一个的崩溃让人无法理解，其他都符合预期，对于这种申请资源失败的情况，目前还没有较好的对策，如果有我会及时补上