CODEWITHSUNDEEP
c#genericsequality
Ahmet Sancar
Ahmet Sancar

Reputation: 65

C# generic types equality operator

From 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);
}

Everything is ok until i tried following, with same method

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

It outputs 'True', s1 and s2 reference different objects in memory even they have same value right? Am i missing something?

Upvotes: 6

Views: 641

Answers (1)

The Vermilion Wizard
The Vermilion Wizard

Reputation: 5395

Strings are interned in .NET, so when you do 

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

they are both pointing to the same object. This is why the MSDN example uses a StringBuilder, this fools the CLR into creating another string object with the same value so that the operator test in the generic method will return false.

Upvotes: 7

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