Question

其中struct在.cpp

中

`

struct {
   float data[Length];
   float convol[Length];
}inkernel;

DeviceSize deviceSize = sizeof(inkernel);

for (uint32_t i = 0;i < Length;i++)
{
    inkernel.convol[i] = 2;
    inkernel.data[i] = 3;

}

status = HPC::bvk::createBuffer(
    info.gpus[0],
    info.device,
    VK_BUFFER_USAGE_STORAGE_BUFFER_BIT,
    VK_MEMORY_PROPERTY_HOST_VISIBLE_BIT | VK_MEMORY_PROPERTY_HOST_COHERENT_BIT,
    deviceSize,
    &inBuffer);
inBuffer.map();
memcpy(inBuffer.mapped, &inkernel, deviceSize);
inBuffer.unmap();`

在着色器中使用两个数组（使用glsl）

`

layout (local_size_x = 4) in;
// Binding 0 : for test
layout(binding = 0) buffer In 
{
   float indata[];
   float cal[];
}inkernel;

layout(binding = 1) buffer Out 
{
   float outdata[];
}outkernel;


void main()
{   
   uint index = gl_GlobalInvocationID.x;
   outkernel.outdata[index] = inkernel.cal[index];
} `

通过计算，我用outkernel.data读取数据，但结果是 3，而不是2 。我不知道为什么inkernel.cal的值被改为 3，而不是2

Answer 1

值不必在同一个ssbo中。

layout(binding = 0) buffer In 
{
   float indata[];
}inkernel1;
layout(binding = 1) buffer In 
{
   float cal[];
}inkernel2;

layout(binding = 2) buffer Out 
{
   float outdata[];
}outkernel;

缓冲区仍然相同，您只需要调整描述符。

如何在comptu着色器（ssbo）的一个缓冲区中使用两个数组？

1 个答案: