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

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

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

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

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

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

cuda本机参数查看的代码

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