Java string concat in stringbuilder call

As far as I know开发者_如何学编程, StringBuilder helps to reduce memory usage by not creating temporary string instances in the string pool during concats. But, what happens if I do sth like this:

StringBuilder sb = new StringBuilder("bu");
sb.append("b"+"u");

Does it compile into

sb.append("b");
sb.append("u");

? Or it depends on optimalization flags? Or I loose the whole benefit if stringbuilders? Or this quetion makes no sense? :)

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It compiles to sb.append("bu"), because the compiler translates the concatenation of multiple String litterals to a single String litteral.

If you had

String a = "a";
sb.append(a + "b");

it would compile it to

String a = "a";
String temp = a + "b"; // useless creation of a string here
sb.append(temp);

So you should prefer

sb.append(a);
sb.append("b");

in this case.

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Since "b" + "u" is an expression which is evaluated at compile time, it will be compiled just as if you had "bu".

 0: new #2; //class StringBuilder
 3: dup
 4: ldc #3; //String bu
 6: invokespecial   #4; //Method StringBuilder."<init>":(String;)V
 9: astore_1
10: aload_1
11: ldc #3; //String bu
13: invokevirtual   #5; // StringBuilder.append:(String;)LStringBuilder;

If you on the other hand had two string variables, this optimization wouldn't kick in:

The following snippet...

StringBuilder sb = new StringBuilder("bu");
String b = "b", u = "u";

sb.append(b + u);

...gets compiled as:

0:  new #2; //class StringBuilder
3:  dup
4:  ldc #3; //String bu
6:  invokespecial   #4; //Method StringBuilder."<init>":(String;)V
9:  astore_1
10: ldc #5; //String b
12: astore_2
13: ldc #6; //String u
15: astore_3
16: aload_1
17: new #2; //class StringBuilder
20: dup
21: invokespecial   #7; //Method StringBuilder."<init>":()V
24: aload_2
25: invokevirtual   #8; //Method StringBuilder.append:(String;)StringBuilder;
28: aload_3
29: invokevirtual   #8; //Method StringBuilder.append:(String;)StringBuilder;
32: invokevirtual   #9; //Method StringBuilder.toString:()String;
35: invokevirtual   #8; //Method StringBuilder.append:(String;)StringBuilder;

I.e. something similar to

StringBuilder sb = new StringBuilder("bu");
String b = "b", u = "u";

StringBuilder temp = new StringBuilder();
temp.append(b);
temp.append(b);
String result = temp.toString();

sb.append(result);

As you can see in line 17-21 an extra StringBuilder is created for the purpose of concatenating a and b. The resulting String of this temporary StringBuilder is then fetched on line 32 and appended to the original StringBuilder on line 35.

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(The bytecode was generated by the javap command which is part of the JDK. Try it out, it's really simple!)

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No, the "b" + "u" will create an immutable b string, an immutable u string, and then create a third immutable bu string that gets passed into the StringBuilder instance.

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WRONG:

StringBuilder sb = new StringBuilder();
sb.append("b"+"u");

CORRECT:

StringBuilder sb = new StringBuilder();
sb.append("b").append("u");

BEST: //Since you knew what's going to be in there already! ;)

StringBuilder sb = new StringBuilder();
sb.append("bu");

:)

EDIT

I guess my answer above is not correct when dealing with literals only...

:/