覆盖GetHashCode（枚举和整数比较）

Question

我想GetHashCode用以下代码覆盖一个方法：

public override int GetHashCode()

{

    var magicNumber1 = 5;

    var magicNumber2 = 3;

    return intType * magicNumber1 + 

          (int)enumType * magicNumber2;

}

GetHashCode当一个值是Integer而第二个是Enum时，这是返回的正确方法吗？

Answer 1

好吧，您当前的实现还可以，但是如果和范围较小，则可以更好。假设（几乎总是）在，并且在范围内，则intTypeenumTypeintType[0..1000000]enumType[0..10]&nbsp; &nbsp;public override int GetHashCode() {&nbsp; &nbsp; &nbsp; return unchecked(intType * 10 + (int)enumType);&nbsp; &nbsp;}将会是一个更好的选择：当前代码中存在的许多哈希冲突在上面的代码中并不存在。例如&nbsp; &nbsp;intType | enumType | old hash | suggested&nbsp; &nbsp;-----------------------------------------&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 5&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;15&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;5&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;3&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;15&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 30&nbsp;&nbsp;编辑：根据您的情况（请查看评论）它不应该超过100假设您没有负值，则可以尝试&nbsp; &nbsp;public override int GetHashCode() {&nbsp; &nbsp; &nbsp;// * 128 == << 7 may be faster than * 100 at some systems&nbsp;&nbsp; &nbsp; &nbsp;return unchecked(intType * 128 + (int)enumType);&nbsp; &nbsp;}并期望完全没有碰撞