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。
找不到一个我完全满意的解决方案(合理的性能,但没有不安全的代码/pinvoke),所以我想出了这个,没有真正的原创,但工作:
/// <summary>
///
/// </summary>
/// <param name="array1"></param>
/// <param name="array2"></param>
/// <param name="bytesToCompare"> 0 means compare entire arrays</param>
/// <returns></returns>
public static bool ArraysEqual(byte[] array1, byte[] array2, int bytesToCompare = 0)
{
if (array1.Length != array2.Length) return false;
var length = (bytesToCompare == 0) ? array1.Length : bytesToCompare;
var tailIdx = length - length % sizeof(Int64);
//check in 8 byte chunks
for (var i = 0; i < tailIdx; i += sizeof(Int64))
{
if (BitConverter.ToInt64(array1, i) != BitConverter.ToInt64(array2, i)) return false;
}
//check the remainder of the array, always shorter than 8 bytes
for (var i = tailIdx; i < length; i++)
{
if (array1[i] != array2[i]) return false;
}
return true;
}
与本页上的其他解决方案相比,性能:
简单循环:19837滴答,1.00
*位收敛器:4886 ticks, 4.06
unsafcompare: 1636 ticks, 12.12
EqualBytesLongUnrolled: 637 tick, 31.09
P/Invoke memcmp: 369 ticks, 53.67
在linqpad上测试,1000000字节的相同数组(最坏的情况),每个数组500次迭代。
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);
}
}