我怎样才能做得快呢?
当然我可以这样做:
static bool ByteArrayCompare(byte[] a1, byte[] a2)
{
if (a1.Length != a2.Length)
return false;
for (int i=0; i<a1.Length; i++)
if (a1[i]!=a2[i])
return false;
return true;
}
但我正在寻找一个BCL函数或一些高度优化的已证明的方法来做到这一点。
java.util.Arrays.equals((sbyte[])(Array)a1, (sbyte[])(Array)a2);
工作得很好，但这似乎不适用于x64。
注意我的快速回答。
当前回答
抱歉，如果你正在寻找一种管理的方式，你已经正确地做了，据我所知，在BCL中没有内置的方法来做这个。
你应该添加一些初始的空检查，然后重用它，就好像它在BCL。
其他回答
我会使用不安全的代码并运行for循环比较Int32指针。
也许您还应该考虑检查数组是否为非空。
抱歉，如果你正在寻找一种管理的方式，你已经正确地做了，据我所知，在BCL中没有内置的方法来做这个。
你应该添加一些初始的空检查，然后重用它，就好像它在BCL。
受到ArekBulski发布的EqualBytesLongUnrolled方法的启发，我确定了一个附加优化的解决方案。在我的实例中，数组中的数组差异往往在数组的尾部附近。在测试中，我发现当这种情况发生在大型数组中时，能够以相反的顺序比较数组元素使这种解决方案比基于memcmp的解决方案获得了巨大的性能提升。下面是解决方案:
public enum CompareDirection { Forward, Backward }
private static unsafe bool UnsafeEquals(byte[] a, byte[] b, CompareDirection direction = CompareDirection.Forward)
{
// returns when a and b are same array or both null
if (a == b) return true;
// if either is null or different lengths, can't be equal
if (a == null || b == null || a.Length != b.Length)
return false;
const int UNROLLED = 16; // count of longs 'unrolled' in optimization
int size = sizeof(long) * UNROLLED; // 128 bytes (min size for 'unrolled' optimization)
int len = a.Length;
int n = len / size; // count of full 128 byte segments
int r = len % size; // count of remaining 'unoptimized' bytes
// pin the arrays and access them via pointers
fixed (byte* pb_a = a, pb_b = b)
{
if (r > 0 && direction == CompareDirection.Backward)
{
byte* pa = pb_a + len - 1;
byte* pb = pb_b + len - 1;
byte* phead = pb_a + len - r;
while(pa >= phead)
{
if (*pa != *pb) return false;
pa--;
pb--;
}
}
if (n > 0)
{
int nOffset = n * size;
if (direction == CompareDirection.Forward)
{
long* pa = (long*)pb_a;
long* pb = (long*)pb_b;
long* ptail = (long*)(pb_a + nOffset);
while (pa < ptail)
{
if (*(pa + 0) != *(pb + 0) || *(pa + 1) != *(pb + 1) ||
*(pa + 2) != *(pb + 2) || *(pa + 3) != *(pb + 3) ||
*(pa + 4) != *(pb + 4) || *(pa + 5) != *(pb + 5) ||
*(pa + 6) != *(pb + 6) || *(pa + 7) != *(pb + 7) ||
*(pa + 8) != *(pb + 8) || *(pa + 9) != *(pb + 9) ||
*(pa + 10) != *(pb + 10) || *(pa + 11) != *(pb + 11) ||
*(pa + 12) != *(pb + 12) || *(pa + 13) != *(pb + 13) ||
*(pa + 14) != *(pb + 14) || *(pa + 15) != *(pb + 15)
)
{
return false;
}
pa += UNROLLED;
pb += UNROLLED;
}
}
else
{
long* pa = (long*)(pb_a + nOffset);
long* pb = (long*)(pb_b + nOffset);
long* phead = (long*)pb_a;
while (phead < pa)
{
if (*(pa - 1) != *(pb - 1) || *(pa - 2) != *(pb - 2) ||
*(pa - 3) != *(pb - 3) || *(pa - 4) != *(pb - 4) ||
*(pa - 5) != *(pb - 5) || *(pa - 6) != *(pb - 6) ||
*(pa - 7) != *(pb - 7) || *(pa - 8) != *(pb - 8) ||
*(pa - 9) != *(pb - 9) || *(pa - 10) != *(pb - 10) ||
*(pa - 11) != *(pb - 11) || *(pa - 12) != *(pb - 12) ||
*(pa - 13) != *(pb - 13) || *(pa - 14) != *(pb - 14) ||
*(pa - 15) != *(pb - 15) || *(pa - 16) != *(pb - 16)
)
{
return false;
}
pa -= UNROLLED;
pb -= UNROLLED;
}
}
}
if (r > 0 && direction == CompareDirection.Forward)
{
byte* pa = pb_a + len - r;
byte* pb = pb_b + len - r;
byte* ptail = pb_a + len;
while(pa < ptail)
{
if (*pa != *pb) return false;
pa++;
pb++;
}
}
}
return true;
}
我使用附带的。net 4.7发布版本做了一些测量，没有附带调试器。我认为人们一直在使用错误的度量，因为如果你关心这里的速度，你所关心的是计算两个字节数组是否相等需要多长时间。即以字节为单位的吞吐量。
StructuralComparison : 4.6 MiB/s
for : 274.5 MiB/s
ToUInt32 : 263.6 MiB/s
ToUInt64 : 474.9 MiB/s
memcmp : 8500.8 MiB/s
正如你所看到的，没有比memcmp更好的方法了，而且它快了几个数量级。简单的for循环是次优选择。我仍然不明白为什么微软不能简单地包含一个缓冲区。比较方法。
[Program.cs]:
using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.InteropServices;
using System.Text;
using System.Threading.Tasks;
namespace memcmp
{
class Program
{
static byte[] TestVector(int size)
{
var data = new byte[size];
using (var rng = new System.Security.Cryptography.RNGCryptoServiceProvider())
{
rng.GetBytes(data);
}
return data;
}
static TimeSpan Measure(string testCase, TimeSpan offset, Action action, bool ignore = false)
{
var t = Stopwatch.StartNew();
var n = 0L;
while (t.Elapsed < TimeSpan.FromSeconds(10))
{
action();
n++;
}
var elapsed = t.Elapsed - offset;
if (!ignore)
{
Console.WriteLine($"{testCase,-16} : {n / elapsed.TotalSeconds,16:0.0} MiB/s");
}
return elapsed;
}
[DllImport("msvcrt.dll", CallingConvention = CallingConvention.Cdecl)]
static extern int memcmp(byte[] b1, byte[] b2, long count);
static void Main(string[] args)
{
// how quickly can we establish if two sequences of bytes are equal?
// note that we are testing the speed of different comparsion methods
var a = TestVector(1024 * 1024); // 1 MiB
var b = (byte[])a.Clone();
// was meant to offset the overhead of everything but copying but my attempt was a horrible mistake... should have reacted sooner due to the initially ridiculous throughput values...
// Measure("offset", new TimeSpan(), () => { return; }, ignore: true);
var offset = TimeZone.Zero
Measure("StructuralComparison", offset, () =>
{
StructuralComparisons.StructuralEqualityComparer.Equals(a, b);
});
Measure("for", offset, () =>
{
for (int i = 0; i < a.Length; i++)
{
if (a[i] != b[i]) break;
}
});
Measure("ToUInt32", offset, () =>
{
for (int i = 0; i < a.Length; i += 4)
{
if (BitConverter.ToUInt32(a, i) != BitConverter.ToUInt32(b, i)) break;
}
});
Measure("ToUInt64", offset, () =>
{
for (int i = 0; i < a.Length; i += 8)
{
if (BitConverter.ToUInt64(a, i) != BitConverter.ToUInt64(b, i)) break;
}
});
Measure("memcmp", offset, () =>
{
memcmp(a, b, a.Length);
});
}
}
}
简单的回答是:
public bool Compare(byte[] b1, byte[] b2)
{
return Encoding.ASCII.GetString(b1) == Encoding.ASCII.GetString(b2);
}
通过这种方式，您可以使用优化的. net字符串比较来进行字节数组比较，而不需要编写不安全的代码。这是它如何在后台完成的:
private unsafe static bool EqualsHelper(String strA, String strB)
{
Contract.Requires(strA != null);
Contract.Requires(strB != null);
Contract.Requires(strA.Length == strB.Length);
int length = strA.Length;
fixed (char* ap = &strA.m_firstChar) fixed (char* bp = &strB.m_firstChar)
{
char* a = ap;
char* b = bp;
// Unroll the loop
#if AMD64
// For the AMD64 bit platform we unroll by 12 and
// check three qwords at a time. This is less code
// than the 32 bit case and is shorter
// pathlength.
while (length >= 12)
{
if (*(long*)a != *(long*)b) return false;
if (*(long*)(a+4) != *(long*)(b+4)) return false;
if (*(long*)(a+8) != *(long*)(b+8)) return false;
a += 12; b += 12; length -= 12;
}
#else
while (length >= 10)
{
if (*(int*)a != *(int*)b) return false;
if (*(int*)(a+2) != *(int*)(b+2)) return false;
if (*(int*)(a+4) != *(int*)(b+4)) return false;
if (*(int*)(a+6) != *(int*)(b+6)) return false;
if (*(int*)(a+8) != *(int*)(b+8)) return false;
a += 10; b += 10; length -= 10;
}
#endif
// This depends on the fact that the String objects are
// always zero terminated and that the terminating zero is not included
// in the length. For odd string sizes, the last compare will include
// the zero terminator.
while (length > 0)
{
if (*(int*)a != *(int*)b) break;
a += 2; b += 2; length -= 2;
}
return (length <= 0);
}
}
