Question

我正在尝试使用clock_gettime函数获取以下代码的运行时。但是当我运行代码时，每次运行时我都会收到0.0000的时间。我也单独输出了开始和停止时间，我收到完全相同的答案。

struct timespec start, stop;
double accum;

if( clock_gettime( CLOCK_REALTIME, &start) == -1 ) {
  perror( "clock gettime" );
  exit( EXIT_FAILURE );
}

int src = 1, final_ret = 0;
for (int t = 0; t < rows - 1; t += pyramid_height)
{
    int temp = src;
    src = final_ret;
    final_ret = temp;

    // Calculate this for the kernel argument...
    int arg0 = MIN(pyramid_height, rows-t-1);
    int theHalo = HALO;

    // Set the kernel arguments.
    clSetKernelArg(cl.kernel(kn), 0,  sizeof(cl_int), (void*) &arg0);
    clSetKernelArg(cl.kernel(kn), 1,  sizeof(cl_mem), (void*) &d_gpuWall);
    clSetKernelArg(cl.kernel(kn), 2,  sizeof(cl_mem), (void*) &d_gpuResult[src]);
    clSetKernelArg(cl.kernel(kn), 3,  sizeof(cl_mem), (void*) &d_gpuResult[final_ret]);
    clSetKernelArg(cl.kernel(kn), 4,  sizeof(cl_int), (void*) &cols);
    clSetKernelArg(cl.kernel(kn), 5,  sizeof(cl_int), (void*) &rows);
    clSetKernelArg(cl.kernel(kn), 6,  sizeof(cl_int), (void*) &t);
    clSetKernelArg(cl.kernel(kn), 7,  sizeof(cl_int), (void*) &borderCols);
    clSetKernelArg(cl.kernel(kn), 8,  sizeof(cl_int), (void*) &theHalo);
    clSetKernelArg(cl.kernel(kn), 9,  sizeof(cl_int) * (cl.localSize()), 0);
    clSetKernelArg(cl.kernel(kn), 10, sizeof(cl_int) * (cl.localSize()), 0);
    clSetKernelArg(cl.kernel(kn), 11, sizeof(cl_mem), (void*) &d_outputBuffer);
    cl.launch(kn);
}

if( clock_gettime( CLOCK_REALTIME, &stop) == -1 ) {
  perror( "clock gettime" );
  exit( EXIT_FAILURE );
}
printf( "%lf\n", stop.tv_sec ); 
 printf( "%lf\n", start.tv_sec );  
accum = ( stop.tv_sec - start.tv_sec )
      + ( stop.tv_nsec - start.tv_nsec )
        / BILLION;
printf( "%lf\n", accum );

非常感谢任何关于我做错事的建议

Answer 1

timespec::tv_nsec是整数类型，因此如果BILLION也是整数类型，那么：

( stop.tv_nsec - start.tv_nsec )
    / BILLION;

将截断为零。如果tv_sec值相同，则会得到零差异。

尝试：

double( stop.tv_nsec - start.tv_nsec )
    / BILLION;

这将使用double类型执行除法。

Clock_gettime（）函数输出错误的时间

1 个答案: