Mutable strings in Java

2020-07-05 06:32发布

问题:

As almost everybody knows strings in Java are immutable. Recently I discovered something that might suggest that it's not always true. Let's try out this code:

System.out.println("-------- BEFORE MODIFICATIONS --------");
String beforeTest = new String("Original");
System.out.println(beforeTest);
java.lang.reflect.Field valueField = String.class.getDeclaredField("value");
valueField.setAccessible(true);
valueField.set("Original", "Modified".toCharArray());
System.out.println("-------- AFTER MODIFICATIONS --------");
System.out.println(beforeTest);
System.out.println("Original");
String test = new String("Original");
System.out.println(test);
String test2 = new String("Original 2");
System.out.println(test2);

the output would be:

-------- BEFORE MODIFICATIONS --------
Original
-------- AFTER MODIFICATIONS --------
Original
Modified
Modified
Original 2

How does this trick work? How does the JVM know which objects should be changed and which not? What mechanism are under the hood of this trick? Why already created beforeTest string was not changed? Does this trick really derogate the strings are immutable principle?

回答1:

String literals are interned into a pool. This means that when you write

String s1 = "Foo";
String s2 = "Foo";
String s3 = new String("Foo");

s1 and s2 refer to the same String object, and s3 refers to another one, backed by another char array.

In your code, you violate String's invariants by modifying the private char array holding the characters of the "Original" String literal instance. But since beforeTest refers to another String instance, it's not modified.

Immutability is achieved by keeping fields private into an object, and not providing any method to modify this private state. By using reflection, you break all the rules of encapsulation, and you can thus violate immutability.