As far as I know, 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? :)
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.
String cd = "cd";, how does the following String get optimized? String abcdef = "a" + "b" + cd + "e" + "f";?? Thanks!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.
(The bytecode was generated by the javap command which is part of the JDK. Try it out, it's really simple!)
String cd = "cd";, are the following two equivalent as well? String abcdef = "a" + "b" + cd + "e" + "f"; <==>String abcdef = "ab" + cd + "ef";`? Thanks!
javap? :-)javac Test.java and then javap -c Test."ab" + cd + "e" + "f"; I tried with different inputs as well, basically only the literals before the variable get optimized. Very interesting.. I guess it might be related to the natural order of '+'..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.
bu string.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...
:/