Simplify byte shuffle to direct scatter writes
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>
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1 file changed
+4
-35
@@ -22,42 +22,11 @@ macro_rules! fp_predict {
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result.resize(start_len + byte_count, 0);
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let output = &mut result[start_len..];
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// Step 1 & 2: Convert to big-endian and shuffle bytes
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// All first bytes, then all second bytes, etc.
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//
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// We process the input in chunks of 4 floats/doubles at a time.
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// For each chunk we shuffle the bytes into a stack buffer, then
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// copy each byte-lane's portion into the output. This turns
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// single-byte pushes into 16-byte (for f32) or 32-byte (for f64)
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// memcpy-like stores that the compiler can vectorize.
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const CHUNK: usize = 4;
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let chunks = num_values / CHUNK;
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let remainder = num_values % CHUNK;
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let mut scratch = [[0u8; CHUNK]; BYTE_SIZE];
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for c in 0..chunks {
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let src = &row[c * CHUNK..][..CHUNK];
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for j in 0..CHUNK {
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let bytes = src[j].to_be_bytes();
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for b in 0..BYTE_SIZE {
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scratch[b][j] = bytes[b];
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}
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}
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// Step 1 & 2: Convert to big-endian and scatter bytes
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for (i, &value) in row.iter().enumerate() {
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let bytes = value.to_be_bytes();
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for b in 0..BYTE_SIZE {
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let dest_offset = b * num_values + c * CHUNK;
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output[dest_offset..dest_offset + CHUNK]
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.copy_from_slice(&scratch[b]);
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}
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}
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// Handle the remaining elements that don't fill a full chunk
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if remainder > 0 {
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let src = &row[chunks * CHUNK..];
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for (j, &value) in src.iter().enumerate() {
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let bytes = value.to_be_bytes();
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for b in 0..BYTE_SIZE {
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output[b * num_values + chunks * CHUNK + j] = bytes[b];
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}
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output[b * num_values + i] = bytes[b];
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}
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}
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