C#的Hash码实现源码

public int GetHashCode(string s)
    {
        int hash = 5381;
        int len = s.Length;

        for (int i = 0; i < len; i++)
        {
            int c = Convert.ToInt32(s[i]);
            hash = ((hash << 5) + hash) ^ c;
        }

        return hash;
    }

System\String.cs

  // Gets a hash code for this string.  If strings A and B are such that A.Equals(B), then
        // they will return the same hash code.
        [System.Security.SecuritySafeCritical]  // auto-generated
        [ReliabilityContract(Consistency.WillNotCorruptState, Cer.MayFail)]
        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);
                }
            }
        }
 

你可能感兴趣的:(.NET)