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Simplify byte shuffle to direct scatter writes

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Replace chunked scratch-buffer approach with direct indexed writes
into the pre-allocated output. Simpler and faster in practice.

Co-Authored-By: Claude <noreply@anthropic.com>
This commit is contained in:
Sergey "Shnatsel" DavidoffandClaude committed 2026-02-27 01:02:41 +00:00
1 parent 2c5fe2c2f5
commit 22046d4272
1 file changed
+4 -35
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src/encoder/colortype.rs
+4 -35
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@@ -22,42 +22,11 @@ macro_rules! fp_predict {
result.resize(start_len + byte_count, 0); result.resize(start_len + byte_count, 0);
let output = &mut result[start_len..]; let output = &mut result[start_len..];
// Step 1 & 2: Convert to big-endian and shuffle bytes // Step 1 & 2: Convert to big-endian and scatter bytes
// All first bytes, then all second bytes, etc. for (i, &value) in row.iter().enumerate() {
// let bytes = value.to_be_bytes();
// We process the input in chunks of 4 floats/doubles at a time.
// For each chunk we shuffle the bytes into a stack buffer, then
// copy each byte-lane's portion into the output. This turns
// single-byte pushes into 16-byte (for f32) or 32-byte (for f64)
// memcpy-like stores that the compiler can vectorize.
const CHUNK: usize = 4;
let chunks = num_values / CHUNK;
let remainder = num_values % CHUNK;
let mut scratch = [[0u8; CHUNK]; BYTE_SIZE];
for c in 0..chunks {
let src = &row[c * CHUNK..][..CHUNK];
for j in 0..CHUNK {
let bytes = src[j].to_be_bytes();
for b in 0..BYTE_SIZE {
scratch[b][j] = bytes[b];
}
}
for b in 0..BYTE_SIZE { for b in 0..BYTE_SIZE {
let dest_offset = b * num_values + c * CHUNK; output[b * num_values + i] = bytes[b];
output[dest_offset..dest_offset + CHUNK]
.copy_from_slice(&scratch[b]);
}
}
// Handle the remaining elements that don't fill a full chunk
if remainder > 0 {
let src = &row[chunks * CHUNK..];
for (j, &value) in src.iter().enumerate() {
let bytes = value.to_be_bytes();
for b in 0..BYTE_SIZE {
output[b * num_values + chunks * CHUNK + j] = bytes[b];
}
} }
} }