Micro Optimization of a 4-bucket histogram of a large array or list

Question

I have a special question. I will try to describe this as accurate as possible.

I am doing a very important \"micro-optimization\". A loop that runs for days at a ti

Answer 1

This is the untested C# version of @PeterCordes answer.

private static Vector128 HsumTranspose( ReadOnlySpan> counts )
{
    var sum01 = Avx2.HorizontalAdd( counts[ 0 ], counts[ 1 ] );
    var sum23 = Avx2.HorizontalAdd( counts[ 2 ], counts[ 3 ] );
    var sum0123 = Avx2.HorizontalAdd( sum01, sum23 );

    var sumHigh = Avx2.ExtractVector128( sum0123, 1 );
    var sumLow = Avx2.ExtractVector128( sum0123, 0 );
    return Sse2.Add( sumHigh, sumLow );
}


private unsafe static int[ ] CountElements( ReadOnlySpan input )
{
    var outputCounts = new int[ 4 ];
    // Four vectors of zeroed counters each vector holds
    // counts for one bucket, to be hsummed at the end.
    Span> counts = stackalloc Vector256[ 4 ]
    {
        Vector256.Zero,
        Vector256.Zero,
        Vector256.Zero,
        Vector256.Zero
    };

    unsafe
    {
        fixed ( int* fixedInput = input )
        {
            var size = input.Length;
            for ( var i = 0; i < size; i += 8 )
            {
                var v = Avx.LoadVector256( &fixedInput[ i ] );
                for ( var val = 0; val < 3; val++ )
                {
                    var match = Avx2.CompareEqual( v, Vector256.Create( val ) );
                    counts[ val ] = Avx2.Subtract( counts[ val ], match );
                }
             }

             Vector128 summedCounts = HsumTranspose( counts );

             fixed ( int* fixedOutputCounts = outputCounts )
                 Sse2.Store( fixedOutputCounts, summedCounts );

             outputCounts[ 3 ] = size - outputCounts[ 0 ] -
                 outputCounts[ 1 ] - outputCounts[ 2 ];

             // TODO: handle the last size%8 input elements; scalar would be easy
            }                
        }            
    }
    return outputCounts;
}