Best practices/performance: mixing StringBuilder.a

Question

I'm trying to understand what the best practice is and why for concatenating string literals and variables for different cases. For instance, if I have code like this

StringBuilder sb = new StringBuilder("AAAAAAAAAAAAA")
    .append(B_String).append("CCCCCCCCCCC").append(D_String)
    .append("EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE")
    .append("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF");

Is this the way to do it? From this post, I noticed that the + operator on Strings creates a new instance of StringBuilder, concatenates the operands, and returns a String conversion, which seems like a lot more work than just calling .append(); so if that's true, then that is out of the question. But what about String.concat()? Is it proper to use .append() for every concatenation? Or just for variables, and literals are okay to append with .concat()?

StringBuilder sb = new StringBuilder("AAAAAAAAAAAAA")
    .append(B_String.concat("CCCCCCCCCCC")).append(D_String
    .concat("EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE")
    .concat("FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF"));

What are the general rules for best practices and performance for going about these situations? Is my assumption correct on + and it should really just not be used?

Answer 1

+ operator

String s = s1 + s2

Behind the scenes this is translated to:

String s = new StringBuilder(s1).append(s2).toString();

Imagine how much extra work it adds if you have s1 + s2 here:

stringBuilder.append(s1 + s2)

instead of:

stringBuilder.append(s1).append(s2)

Multiple strings with +

Worth to note that:

String s = s1 + s2 + s3 + ... +sN

is translated to:

String s = new StringBuilder(s1).append(s2).append(s3)...apend(sN).toString();

concat()

String s = s1.concat(s2);

String creates char[] array that can fit both s1 and s2. Copies s1 and s2 contents to this new array. Actually requires less work then + operator.

StringBuilder.append()

Maintains an internal char[] array that grows when needed. No extra char[] is created if the internal one is sufficiently big.

stringBuilder.append(s1.concat(s2))

is also performing poorly because s1.concat(s2) creates an extra char[] array and copies s1 and s2 to it just to copy that new array contents to internal StringBuilder char[].

That being said you should use append() all the time and append raw strings (your first code snippet is correct).

Answer 2

The compilier optimize the + concatenation.

So

int a = 1;
String s = "Hello " + a;

is transformed into

new StringBuilder().append("Hello ").append(1).toString();

There an excellent topic here explaining why you should use the + operator.

Answer 3

Use + operator is best practice, it is also simple and readable.

The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings. String concatenation is implemented through the StringBuilder(or StringBuffer) class and its append method.

Offical document: https://docs.oracle.com/javase/8/docs/api/java/lang/String.html

Answer 4

in byte code level there is not different and we are not compromising effeciancy there. In case of executing byte code level, it must go through non-inline operator overloading method for + by calling append. then in assembly language level (Java is written in C and C produces assemblies similar to assembly, there will be extra register call to store + method call in the stack and there will additional push. (in reality, cross-compiler might optimise + operator call, in that case making it no difference with efficiancy.)

It is a good practice to have one way to increase the readability. :)

Answer 5

You should always use append.

concat create a new String so it's pretty like + I think.

If you concat or use + with 2 final String the JVM can make optimisation so it's the same as doing append in this case.

Answer 6

Optimization is done automatically by the compiler.

The Java2 compiler will automatically convert the following:

String s = s1 + s2; 

to

String s = (new StringBuffer()).append(s1).append(s2).toString();

Taken straight from the Java Best Practices on Oracles website.