更新时间:2023-09-24 16:26:46
一定要得到的当你有这样的问题,参考源源$ C $ C 。还有很多更给它比你可以从反编译器看到的。挑选一个符合你的preferred .NET的目标,该方法已经改变版本之间有很多。我就重现了.NET 4.5的版本在这里,从Source.NET 4.5 \ 4.6.0.0 \网络检索\ CLR的\ src \ BCL \ SYSTEM \ String.cs \ 604718 \ String.cs
Be sure to obtain the Reference Source source code when you have questions like this. There's a lot more to it than what you can see from a decompiler. Pick the one that matches your preferred .NET target, the method has changed a great deal between versions. I'll just reproduce the .NET 4.5 version of it here, retrieved from Source.NET 4.5\4.6.0.0\net\clr\src\BCL\System\String.cs\604718\String.cs
public override int GetHashCode() {
#if FEATURE_RANDOMIZED_STRING_HASHING
if(HashHelpers.s_UseRandomizedStringHashing)
{
return InternalMarvin32HashString(this, this.Length, 0);
}
#endif // FEATURE_RANDOMIZED_STRING_HASHING
unsafe {
fixed (char *src = this) {
Contract.Assert(src[this.Length] == '\0', "src[this.Length] == '\\0'");
Contract.Assert( ((int)src)%4 == 0, "Managed string should start at 4 bytes boundary");
#if WIN32
int hash1 = (5381<<16) + 5381;
#else
int hash1 = 5381;
#endif
int hash2 = hash1;
#if WIN32
// 32 bit machines.
int* pint = (int *)src;
int len = this.Length;
while (len > 2)
{
hash1 = ((hash1 << 5) + hash1 + (hash1 >> 27)) ^ pint[0];
hash2 = ((hash2 << 5) + hash2 + (hash2 >> 27)) ^ pint[1];
pint += 2;
len -= 4;
}
if (len > 0)
{
hash1 = ((hash1 << 5) + hash1 + (hash1 >> 27)) ^ pint[0];
}
#else
int c;
char *s = src;
while ((c = s[0]) != 0) {
hash1 = ((hash1 << 5) + hash1) ^ c;
c = s[1];
if (c == 0)
break;
hash2 = ((hash2 << 5) + hash2) ^ c;
s += 2;
}
#endif
#if DEBUG
// We want to ensure we can change our hash function daily.
// This is perfectly fine as long as you don't persist the
// value from GetHashCode to disk or count on String A
// hashing before string B. Those are bugs in your code.
hash1 ^= ThisAssembly.DailyBuildNumber;
#endif
return hash1 + (hash2 * 1566083941);
}
}
}
这是比你讨价还价的可能更多,我会注释codeA位:
This is possibly more than you bargained for, I'll annotate the code a bit: