The discussion clarifies the difference between creating strings in Java using new String("test") and using string literals like "test". The former creates a new string instance each time, resulting in different memory references even for equal strings, while the latter utilizes string interning, allowing multiple references to the same instance for identical literals. This leads to more efficient memory usage. Developers are advised to prefer string literals for initialization to optimize performance.
Java developers, software engineers, and anyone interested in optimizing string handling and memory usage in Java applications.
String a = new String("test");
String b = new String("test");
assert a != b;
assert a.equals(b);String a = "test";
String b = "test";
assert a == b;
assert a.equals(b);Filip Larsen said:Assuming that you meant String("test") in the first case, it will create a new string instance. Thus, if you have two such strings they would always be different instances even if the strings themselves are equal, that is
Code:String a = new String("test"); String b = new String("test"); assert a != b; assert a.equals(b);
In the second case you are referencing a so-called interned string instance (see [1] and [2]). This means that String a = "test"; is equivalent to String a = new String("test").intern(); which means that if you have two equal text literals (or other interned strings) they will reference the same instance, that is
Code:String a = "test"; String b = "test"; assert a == b; assert a.equals(b);
In most circumstances you will want to use the later approach to initialize strings from string literals in Java as multiple uses of the same literal will result in only one string instance being used.
[1] http://en.wikipedia.org/wiki/String_interning
[2] http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#intern()