C# 泛型类型相等运算符

Question

来自https://msdn.microsoft.com/en-us/library/d5x73970.aspx

When applying the where T : class constraint, avoid the == and != operators on the type parameter because these operators will test for reference identity only, not for value equality. This is the case even if these operators are overloaded in a type that is used as an argument. The following code illustrates this point; the output is false even though the String class overloads the == operator.

public static void OpTest<T>(T s, T t) where T : class
{
    System.Console.WriteLine(s == t);
}

static void Main()
{
    string s1 = "target";
    System.Text.StringBuilder sb = new System.Text.StringBuilder("target");
    string s2 = sb.ToString();
    OpTest<string>(s1, s2);
}

一切正常，直到我用同样的方法尝试跟随

static void Main()
{
    string s1 = "target";
    string s2 = "target";
    OpTest<string>(s1, s2);
}

它输出 'True'，s1 和 s2 引用内存中的不同对象，即使它们具有相同的值，对吗？我错过了什么吗？

Answer 1

字符串在 .NET 中是固定的，所以当您这样做时

string s1 = "target";
string s2 = "target";

它们都指向同一个对象。这就是 MSDN 示例使用 StringBuilder 的原因，这会欺骗 CLR 创建另一个具有相同值的字符串对象，以便泛型方法中的运算符测试将 return false.

C# 泛型类型相等运算符

C# generic types equality operator

c#

generics

equality