You're on the right track by understanding the immutability of the String class.
Based on [1] and [2], here are some cases where each type of implementation is recommended:
1. Simple String Concatenation
String answer = firstPart + "." + secondPart;
This is syntactic sugar for
String answer = new StringBuilder(firstPart).append("."). append(secondPart).toString();
This is actually quite performant and is the recommended approach for simple string concatenation [1].
2. Stepwise Construction
String answer = firstPart;
answer += ".";
answer += secondPart;
Under the hood, this translates to
String answer = new StringBuilder(firstPart).toString();
answer = new StringBuilder(answer).append(".").toString();
answer = new StringBuilder(answer).append(secondPart).toString();
This creates a temporary StringBuilder and intermediate String objects which are inefficient [1]. Especially if the intermediate results are not used.
Use StringBuilder in this case.
3. For Loop Construction and Scaling For Larger Collections
String result = "";
for(int i = 0; i < numItems(); i++)
result += lineItem(i);
return result;
The above code is O(n^2), where n is number of strings. This is due to the immutability of the String class and due to the the fact that when concatenating two strings, the contents of both are copied [2].
So it may be fine for a few fixed length items, but it will not scale.
In such cases, use StringBuilder.
StringBuilder sb = new StringBuilder(numItems() * LINE_SIZE);
for(int i = 0; i < numItems(); i++)
sb.append(lineItem(i));
return b.toString();
This code is O(n) time, where n is number of items or strings. So as the number of strings gets larger, you will see the difference in performance [2].
This code pre-allocates an array in the initialization of StringBuilder, but even if a default size array is used, it will be significantly faster than the previous code for a large number of items [2].
Summary
Use string concatenation if you are concatenating only a few strings or if performance is not of importance (i.e. a demonstration/toy-application). Otherwise, use StringBuilder or consider processing the string as a character array [2].
References:
[1] Java Performance: The Definitive Guide by Scott Oaks: Link
[2] Effective Java 3rd Edition by Joshua Bloch: Link
Answer from mukundvemuri on Stack OverflowList out some Java String builder methods
What is string builder in Java?
What are some of the benefits of StringBuilder?
Videos
You're on the right track by understanding the immutability of the String class.
Based on [1] and [2], here are some cases where each type of implementation is recommended:
1. Simple String Concatenation
String answer = firstPart + "." + secondPart;
This is syntactic sugar for
String answer = new StringBuilder(firstPart).append("."). append(secondPart).toString();
This is actually quite performant and is the recommended approach for simple string concatenation [1].
2. Stepwise Construction
String answer = firstPart;
answer += ".";
answer += secondPart;
Under the hood, this translates to
String answer = new StringBuilder(firstPart).toString();
answer = new StringBuilder(answer).append(".").toString();
answer = new StringBuilder(answer).append(secondPart).toString();
This creates a temporary StringBuilder and intermediate String objects which are inefficient [1]. Especially if the intermediate results are not used.
Use StringBuilder in this case.
3. For Loop Construction and Scaling For Larger Collections
String result = "";
for(int i = 0; i < numItems(); i++)
result += lineItem(i);
return result;
The above code is O(n^2), where n is number of strings. This is due to the immutability of the String class and due to the the fact that when concatenating two strings, the contents of both are copied [2].
So it may be fine for a few fixed length items, but it will not scale.
In such cases, use StringBuilder.
StringBuilder sb = new StringBuilder(numItems() * LINE_SIZE);
for(int i = 0; i < numItems(); i++)
sb.append(lineItem(i));
return b.toString();
This code is O(n) time, where n is number of items or strings. So as the number of strings gets larger, you will see the difference in performance [2].
This code pre-allocates an array in the initialization of StringBuilder, but even if a default size array is used, it will be significantly faster than the previous code for a large number of items [2].
Summary
Use string concatenation if you are concatenating only a few strings or if performance is not of importance (i.e. a demonstration/toy-application). Otherwise, use StringBuilder or consider processing the string as a character array [2].
References:
[1] Java Performance: The Definitive Guide by Scott Oaks: Link
[2] Effective Java 3rd Edition by Joshua Bloch: Link
You cannot change the original string because it is immutable therefore having String s = ""; every operation like
s += "something";
will create and reassign new object (probably it will also add a little bit of work for GC in near future). On he other hand modifying StringBuilder is (usually) not creating new object (indeed it is happening just once at the very end when calling toString() method on builder instance)
Because of this it is common to use StringBuilder when you are modifying string many many times (for example in some long loops).
Still it is common error to overuse StringBuilder - it may be example of premature optimization
Read also:
- Is it better to reuse a StringBuilder in a loop?
Going through firecode.io and saw this solution (to replace spaces with a certain string)
public static String replace(String a, String b) {
String ans = "";
for (char c: a.toCharArray() ){
if (c == ' '){ ans += b; }
else { ans += c; }
}
return ans;
}A user commented and said:
Use StringBuilder instead - it's more efficient! String does not allow appending. Each method you invoke on a String creates a new object and returns it. This is because String is immutable - it cannot change its internal state. On the other hand StringBuilder is mutable. When you call append(..) it alters the internal char array, rather than creating a new string object.
I'm curious about this and wanted to ask (I'm new to Java):
-
If a traditional 'String' is immutable, why am I able to use it in a "+=" operation to append a char at the end?
-
So when you use the "+=" operation, the computer instantiates a brand new String object each and every time you do it? So effectively I'm creating/destroying multiple String objects over and over again with every "+="?
-
Coming from C++, I see a lot of the "+=" operation when working with strings. Is it the same situation in C++ as it is expressed here in Java?
So far can't really tell the benefit of using it over Strings.