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String interning?

The second ReferenceEquals call returns false. Why isn't the string in s4 interned? (I don't care about the advantages of StringBuilder over string concatenation.)

string s1 = "tom";
string s2 = "tom";


Console.Write(object.ReferenceEquals(s2, s1)); //true

string s3 = "tom";
string s4 = "to";
s4 += "m";

Console.Write(object.ReferenceEquals(s3, s4)); //false

When I do String.Intern(s4);, I still get false.

Here, both s3 and s4 are interned but their references are not equal?

string s3 = "tom";
string s4 = "to";
s4 += "m";
String.Intern(s4);

Console.WriteLine(s3 == s4); //true
Console.WriteLine(object.ReferenceEquals(s3, s4)); //false
Console.WriteLine(string.IsInterned(s3) != null);  //true (s3 is interned)
Console.WriteLine(string.IsInterned(s4) != null);  //true (s4 is interned)
like image 323
rkrauter Avatar asked Apr 24 '10 23:04

rkrauter


1 Answers

The string in s4 is interned. However, when you execute s4 += "m";, you have created a new string that will not be interned as its value is not a string literal but the result of a string concatenation operation. As a result, s3 and s4 are two different string instances in two different memory locations.

For more information on string interning, look here, specifically at the last example. When you do String.Intern(s4), you are indeed interning the string, but you are still not performing a reference equality test between those two interned strings. The String.Intern method returns the interned string, so you would need to do this:

string s1 = "tom";
string s2 = "tom";

Console.Write(object.ReferenceEquals(s2, s1)); //true 

string s3 = "tom";
string s4 = "to";
s4 += "m";

Console.Write(object.ReferenceEquals(s3, s4)); //false

string s5 = String.Intern(s4);

Console.Write(object.ReferenceEquals(s3, s5)); //true
like image 66
Scott Dorman Avatar answered Nov 02 '22 07:11

Scott Dorman