multithreading - Gstreamer缓冲区与CUDA转换

Question

我想使用cuda转换(尝试将红色 channel 设置为0)gstreamer缓冲区。有一段代码:

__global__ void transform( guint8 *data, int width ){
  uint x = (blockIdx.x * blockDim.x) + threadIdx.x;
  uint y = (blockIdx.y * blockDim.y) + threadIdx.y;
  uint pixPos = (y * width + x) * 4;
  data[pixPos + 2] = 0; // BGRA format
}

void simple_transform( guint8 *data, int width, int height ){
  guint8 *d_data;
  guint8 size = width * height * 4;

  checkCudaErrors( cudaMalloc( (void**)&d_data, size ) );
  // copy original buffer into device
  checkCudaErrors( cudaMemcpy( d_data, data, size, cudaMemcpyHostToDevice ) );

  dim3 threads = dim3(8, 8);
  dim3 blocks = dim3(width / threads.x, height / threads.y);

  // execute kernel
  transform<<< blocks, threads >>>( d_data, width );

  // move back converted data to original buffer
  checkCudaErrors( cudaMemcpy( data, d_data, size, cudaMemcpyDeviceToHost ) );

  cudaFree( d_data );
}

问题是视频显示没有任何变化。我想看蓝绿色的图片，但是看不到。我的错误在哪里？

Answer 1

您正在将图像的总大小分配给guint8类型的变量

guint8 size = width * height * 4;

最多只能容纳255个值。必须将大小分配给较大数据类型的变量(例如 size_t)。

size_t size = width * height * 4;