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我有示例代码,但它完全遗漏了我的(void *)should_be!如何在OpenCL中表示图像
我设置一个cl_image_desc,cl_image_format,缓存,产地,以及区域:
cl_image_desc desc;
desc.image_type = CL_MEM_OBJECT_IMAGE2D;
desc.image_width = width;
desc.image_height = height;
desc.image_depth = 0;
desc.image_array_size = 0;
desc.image_row_pitch = 0;
desc.image_slice_pitch = 0;
desc.num_mip_levels = 0;
desc.num_samples = 0;
desc.buffer = NULL;
cl_image_format format;
format.image_channel_order = CL_R;
format.image_channel_data_type = CL_FLOAT;
cl_mem bufferSourceImage = clCreateImage(context, CL_MEM_READ_ONLY, &format, &desc, NULL, NULL);
size_t origin[3] = {0, 0, 0};
size_t region[3] = {width, height,1};
在接下来的片段sourceImage是一个空指针,以我的形象。但是我的形象是什么?对于每个像素有r,g,b,a,x和y值。
clEnqueueWriteImage(queue, bufferSourceImage, CL_TRUE, origin, region, 0, 0, sourceImage, 0, NULL, NULL);
如何将我的图像(一堆(r,g,b,a,x,y))转换为合适的数组?
这是他们所提供的内核:
__kernel void convolution(__read_only image2d_t sourceImage, __write_only image2d_t outputImage, int rows, int cols, __constant float* filter, int filterWidth, sampler_t sampler)
{
int column = get_global_id(0);
int row = get_global_id(1);
int halfWidth = (int)(filterWidth/2);
float4 sum = {0.0f, 0.0f, 0.0f, 0.0f};
int filterIdx = 0;
int2 coords;
for(int i = -halfWidth; i <= halfWidth; i++)
{
coords.y = row + i;
for(int i2 = -halfWidth; i2 <= halfWidth; i2++)
{
coords.x = column + i2;
float4 pixel;
pixel = read_imagef(sourceImage, sampler, coords);
sum.x += pixel.x * filter[filterIdx++];
}
}
if(myRow < rows && myCol < cols)
{
coords.x = column;
coords.y = row;
write_imagef(outputImage, coords, sum);
}
}
好了,所以根据这个表格http://www.khronos.org/registry/cl/sdk/1.0/docs/man/xhtml/read_imagef2d.html会我(将图像转换为数组的术语是什么?)将图像转换为看起来像{{r,g,b,1},{r,g,b,1}等}的数组,其中每个元素被映射到图像的行* w + col? – user1873073
对此没有单一术语。我猜你读了一张图片,你的频道长8位。 CL_R表示三个颜色通道中只有一个通道。示例代码来自哪里? –
'clCreateImage'为您创建了正确的维度矢量数组,您必须填写R通道。 –