Vulkan：统一缓冲区的奇怪性能

Question

One of the inputs of my fragment shader is an array of 5 structures. The shader computes a color based on each of the 5 structures. In the end, these 5 colors are summed together to produce the final output. The total size of the array is 1440 bytes. To accommodate the alignment of the uniform buffer, the size of the uniform buffer changes to 1920 bytes.

1- If I define the array of 5 structures as a uniform buffer array, the rendering takes 5ms (measured by Nsight Graphics). The uniform buffer's memory property is 'VK_MEMORY_PROPERTY_HOST_VISIBLE_BIT | VK_MEMORY_PROPERTY_HOST_COHERENT_BIT'. The uniform buffer in glsl is defined as follows

layout(set=0,binding=0) uniform UniformStruct { A a; } us[];

layout(location=0) out vec4 c;
    
void main() 
{
    vec4 col = vec4(0); 
    for (int i = 0; i < 5; i++)   
      col += func(us[nonuniformEXT(i)]);
    c = col;
}

Besides, I'm using 'GL_EXT_nonuniform_qualifier' extension to access the uniform buffer array. This seems the most straightforward way for me but there are alternative implementations.

2- I can split the rendering from one vkCmdDraw to five vkCmdDraw, change the framebuffer's blend mode from overwriting to addition and define a uniform buffer instead of a uniform buffer array in the fragment shader. On the CPU side, I change the descriptor type from UNIFORM_BUFFER to UNIFORM_BUFFER_DYNAMICS. Before each vkCmdDraw, I bind the dynamic uniform buffer and the corresponding offsets. In the fragment shader, the for loop is removed. Although it seems that it should be slower than the first method, it is surprisingly much faster than the first method. The rendering only takes 2ms total for 5 draws.

3- If I define the array of 5 structures as a storage buffer and do one vkCmdDraw, the rendering takes only 1.4ms. In other words, if I change the array from the uniform buffer array to storage buffer but keep anything else the same as 1, it becomes faster.

4- If I define the array of 5 structures as a global constant in the glsl and do one vkCmdDraw, the rendering takes only 0.5ms.

In my opinion, 4 should be the fastest way, which is true in the test. Then 1 should be the next. Both 2 and 3 should be slower than 1. However, Neither 2 or 3 is slower than 1. In contrast, they are much faster than 1. Any ideas why using uniform buffer array slows down the rendering? Is it because it is a host visible buffer?

Answer 1

When it comes to UBOs, there are two kinds of hardware: the kind where UBOs are specialized hardware and the kind where they aren't. For GPUs where UBOs are not specialized hardware, a UBO is really just a readonly SSBO. You can usually tell the difference because hardware where UBOs are specialized will have different size limits on them from those of SSBOs.

For specialized hardware-based UBOs (which NVIDIA still uses, if I recall correctly), each UBO represents an upload from memory into a big block of constant data that all invocations of a particular shader stage can access.

For this kind of hardware, an array of UBOs is basically creating an array out of segments of this block of constant data. And some hardware has multiple blocks of constant data, so indexing then with non-constant expressions is tricky. This is why non-constant access to such indices is an optional feature of Vulkan.

By contrast, a UBO which contains an array is just one big UBO. It's special only in how big it is. Indexing through an array within a UBO is no different from indexing any array. There are no special rules with regard to the uniformity of the index of such accesses.

So stop using an array of UBOs and just use a single UBO which contains an array of data:

layout(set=0,binding=0) uniform UniformStruct { A a[5]; } us;

It'll also avoid additional padding due to alignment, additional descriptors, additional buffers, etc.

However, you might also speed things up by not lying to Vulkan. The expression nonuniformEXT(i) states that the expression i is not dynamically uniform . This is incorrect. Every shader invocation that executes this loop will generate i expressions that have values from 0 to 4. Every dynamic instance of the expression i for any invocation will have the same value at that place in the code as every other.

Therefore i is dynamically uniform, so telling Vulkan that it isn't is not helpful.