Why does Java's concat() method not do anything?

问题

This code:

String s = \"TEST\";
String s2 = s.trim();

s.concat(\"ING\");
System.out.println(\"S = \"+s);
System.out.println(\"S2 = \"+s2);

results in this output:

S = TEST
S2 = TEST
BUILD SUCCESSFUL (total time: 0 seconds)

Why are \"TEST\" and \"ING\" not concatenated together?

回答1:

a String is immutable, meaning you cannot change a String in Java. concat() returns a new, concatenated, string.

String s = "TEST";
String s2 = s.trim();
String s3 = s.concat("ING");

System.out.println("S = "+s);
System.out.println("S2 = "+s2);
System.out.println("S3 = "+s3);

回答2:

Because String is immutable - class String does not contain methods that change the content of the String object itself. The concat() method returns a new String that contains the result of the operation. Instead of this:

s.concat("ING");

Try this:

s = s.concat("ING");

回答3:

I think what you want to do is:

s2 = s.concat("ING");

The concat function does not change the string s, it just returns s with the argument appended.

回答4:

concat returns a string, so you are basically calling concat without storing what it returns. Try this:

String s = "Hello";
String str = s.concat(" World");
System.out.println(s);
System.out.println(str);

Should print:

Hello
Hello World

回答5:

.concat() method returns new concatenated value. there you did s.concat("ING")--> this gives return value "Testing". But you have to receive it in declared variable.

If you did s=s.concat("ING"); then the value of "s" will change into "TESTING"

回答6:

As String is immutable when concat() is called on any String obj as

String s ="abc"; new String obj is created in String pool ref by s

s = s.concat("def"); here one more String obj with val abcdef is created and referenced by s

来源：https://stackoverflow.com/questions/2818796/why-does-javas-concat-method-not-do-anything