在.Net 7中Math.Round性能优化
前言
在.Net源码看到一个issue,关于Math.Round的性能优化.主要是在使用Round通过指定舍入模式(MidpointRounding.ToEven).- Optimize Math.Round(x, MidpointRounding.AwayFromZero/ToEven) (#64016)
此次代码优化,没有JIT相关的代码.
测试代码
namespace net6perf.Maths
{
[DisassemblyDiagnoser(printSource: true, maxDepth: 3)]
[MemoryDiagnoser]
public class RoundTest
{
[Params(1.5, 2.5, 3.5)]
public double Value { get; set; }
[Benchmark]
public double ToEvenTest() => Math.Round(Value, MidpointRounding.ToEven);
[Benchmark]
public double AwayFromZeroTest() => Math.Round(Value, MidpointRounding.AwayFromZero);
}
}测试结果:
发现ToEven模式性能提升比较明显,而AwayFromZero模式没有太大的变化.在去看源码,ToEven模式调用Round方法(带有Intrinsic标记,指令优化).
[Intrinsic]
public static double Round(double a)
{
// ************************************************************************************
// IMPORTANT: Do not change this implementation without also updating MathF.Round(float),
// FloatingPointUtils::round(double), and FloatingPointUtils::round(float)
// ************************************************************************************
// This is based on the 'Berkeley SoftFloat Release 3e' algorithm
ulong bits = BitConverter.DoubleToUInt64Bits(a);
int exponent = double.ExtractExponentFromBits(bits);
if (exponent <= 0x03FE)
{
if ((bits << 1) == 0)
{
// Exactly +/- zero should return the original value
return a;
}
// Any value less than or equal to 0.5 will always round to exactly zero
// and any value greater than 0.5 will always round to exactly one. However,
// we need to preserve the original sign for IEEE compliance.
double result = ((exponent == 0x03FE) && (double.ExtractSignificandFromBits(bits) != 0)) ? 1.0 : 0.0;
return CopySign(result, a);
}
if (exponent >= 0x0433)
{
// Any value greater than or equal to 2^52 cannot have a fractional part,
// So it will always round to exactly itself.
return a;
}
// The absolute value should be greater than or equal to 1.0 and less than 2^52
Debug.Assert((0x03FF <= exponent) && (exponent <= 0x0432));
// Determine the last bit that represents the integral portion of the value
// and the bits representing the fractional portion
ulong lastBitMask = 1UL << (0x0433 - exponent);
ulong roundBitsMask = lastBitMask - 1;
// Increment the first fractional bit, which represents the midpoint between
// two integral values in the current window.
bits += lastBitMask >> 1;
if ((bits & roundBitsMask) == 0)
{
// If that overflowed and the rest of the fractional bits are zero
// then we were exactly x.5 and we want to round to the even result
bits &= ~lastBitMask;
}
else
{
// Otherwise, we just want to strip the fractional bits off, truncating
// to the current integer value.
bits &= ~roundBitsMask;
}
return BitConverter.UInt64BitsToDouble(bits);
}为此调整测试代码:
namespace net6perf.Maths
{
[DisassemblyDiagnoser(printSource: true, maxDepth: 3)]
[MemoryDiagnoser]
public class RoundTest
{
[Params(1.5, 2.5, 3.5)]
public double Value { get; set; }
[Benchmark]
public double ToEvenTest() => Math.Round(Value, MidpointRounding.ToEven);
[Benchmark]
public double AwayFromZeroTest() => Math.Round(Value, MidpointRounding.AwayFromZero);
[Benchmark(Baseline = true)]
public double RoundDefault() => Math.Round(Value);
}
}看生成汇编代码:
; net6perf.Maths.RoundTest.ToEvenTest()
vzeroupper
vroundsd xmm0,xmm0,qword ptr [rcx+8],4
ret
; Total bytes of code 11
; net6perf.Maths.RoundTest.RoundDefault()
vzeroupper
vroundsd xmm0,xmm0,qword ptr [rcx+8],4
ret
; Total bytes of code 11AwayFromZero没有做优化,生成的汇编代码太长,所以就没有展示.
秋风
2022-03-28