问题
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)
回答1:
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
回答2:
Strings are immutable. This means their contents can't be changed.
When you do s4 += "m"; internally, the CLR copies the string to another location in memory which contains the original string and the appended part.
See MSDN string reference.
回答3:
Source: https://blogs.msdn.microsoft.com/ericlippert/2009/09/28/string-interning-and-string-empty/
String interning is an optimization technique by the compiler. If you have two identical string literals in one compilation unit then the code generated ensures that there is only one string object created for all the instance of that literal(characters enclosed in double quotes) within the assembly.
I am from C# background, so i can explain by giving a example from that:
object obj = "Int32";
string str1 = "Int32";
string str2 = typeof(int).Name;
output of the following comparisons:
Console.WriteLine(obj == str1); // true
Console.WriteLine(str1 == str2); // true
Console.WriteLine(obj == str2); // false !?
Note1:Objects are compared by reference.
Note2:typeof(int).Name is evaluated by reflection method so it does not gets evaluated at compile time. Here these comparisons are made at compile time.
Analysis of the Results: 1) true because they both contain same literal and so the code generated will have only one object referencing "Int32". See Note 1.
2) true because the content of both the value is checked which is same.
3) FALSE because str2 and obj does not have the same literal. See Note 2.
回答4:
First of all, everything written so far about immutable strings is correct. But there are some important things which are not written. The code
string s1 = "tom";
string s2 = "tom";
Console.Write(object.ReferenceEquals(s2, s1)); //true
display really "True", but only because of some small compiler optimization or like here because CLR ignore C# compiler attributes (see "CLR via C#" book) and place only one string "tom" in the heap.
Second you can fix the situation with following lines:
s3 = String.Intern(s3);
s4 = String.Intern(s4);
Console.Write (object.ReferenceEquals (s3, s4)); //true
Function String.Intern calculates a hash code of the string and search for the same hash in the internal hash table. Because it find this, it returns back the reference to already existing String object. If the string doesn't exist in the internal hash table, a copy of the string is made and the hash computed. The garbage collector doesn't free memory for the string, because it is referenced by the hash table.
回答5:
In C#, each string is a distinct object, and cannot be edited. You are creating references to them, but each string is distinct. The behaviour is consistent and easy to understand.
Might I suggest examining the StringBuilder class for manipulating strings without creating new instances? It should be sufficient for anything you want to do with strings.
回答6:
When comparing two objects, not strings, the string equality operator is not called since it is static method without polymorphism.
Here is a test:
static void Test()
{
object o1 = "a";
object o2 = new string("a".ToCharArray());
string o3 = "a";
string o4 = new string("a".ToCharArray());
object o5 = "a"; // Compiler optimization addr(o5) = addr(o6)
object o6 = "a";
string o7 = "a"; // Compiler optimization addr(o7) = addr(o8)
string o8 = "a";
Console.WriteLine("Enter same text 4 times:");
object o9 = Console.ReadLine();
object o10 = Console.ReadLine();
string o11 = Console.ReadLine();
string o12 = Console.ReadLine();
Console.WriteLine("object arr o1 == o2 ? " + ( o1 == o2 ).ToString());
Console.WriteLine("string arr o3 == o4 ? " + ( o3 == o4 ).ToString());
Console.WriteLine("object const o5 == o6 ? " + ( o5 == o6 ).ToString());
Console.WriteLine("string const o7 == o8 ? " + ( o7 == o8 ).ToString());
Console.WriteLine("object cnsl o9 == o10 ? " + ( o9 == o10 ).ToString());
Console.WriteLine("string cnsl o11 == o12 ? " + ( o11 == o12 ).ToString());
Console.WriteLine("o1.Equals(o2) ? " + o1.Equals(o2).ToString());
Console.WriteLine("o3.Equals(o4) ? " + o3.Equals(o4).ToString());
Console.WriteLine("o5.Equals(o6) ? " + o5.Equals(o6).ToString());
Console.WriteLine("o7.Equals(o8) ? " + o7.Equals(o8).ToString());
Console.WriteLine("o9.Equals(o10) ? " + o9.Equals(o11).ToString());
Console.WriteLine("o11.Equals(o12) ? " + o11.Equals(o12).ToString());
}
Results:
object arr o1 == o2 ? False
string arr o3 == o4 ? True
object const o5 == o6 ? True
string const o7 == o8 ? True
object cnsl o9 == o10 ? False
string cnsl o11 == o12 ? True
o1.Equals(o2) ? True
o3.Equals(o4) ? True
o5.Equals(o6) ? True
o7.Equals(o8) ? True
o9.Equals(o10) ? True
o11.Equals(o12) ? True
https://referencesource.microsoft.com/#mscorlib/system/string.cs
来源：https://stackoverflow.com/questions/2706607/string-interning