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java I thought strings were immutable

I was always told strings in java are immutable, unless your going to use the string builder class or string writter class.
Take a look at this practice question I found online,
Given a string and a non-negative int n, return a larger string that is n copies of the original string.
stringTimes("Hi", 2) → "HiHi"
stringTimes("Hi", 3) → "HiHiHi"
stringTimes("Hi", 1) → "Hi"
and the solution came out to be
Solution:
public String stringTimes(String str, int n) {
String result = "";
for (int i=0; i<n; i++) {
result = result + str; // could use += here
}
return result;
}
As you see in the solution our 3rd line assigns the string , then we change it in our for loop. This makes no sense to me! (I answered the question in another nooby way ) Once I saw this solution I knew I had to ask you guys.
Thoughts? I know im not that great at programming but I haven't seen this type of example here before, so I thought I'd share.

The trick to understanding what's going on is the line below:
result = result + str;
or its equivalent
result += str;
Java compiler performs a trick on this syntax - behind the scene, it generates the following code:
result = result.concat(str);
Variable result participates in this expression twice - once as the original string on which concat method is called, and once as the target of an assignment. The assignment does not mutate the original string, it replaces the entire String object with a new immutable one provided by concat.
Perhaps it would be easier to see if we introduce an additional String variable into the code:
String temp = result.concat(str);
result = temp;
Once the first line has executed, you have two String objects - temp, which is "HiHi", and result, which is still "Hi". When the second line is executed, result gets replaced with temp, acquiring a new value of "HiHi".

If you use Eclipse, you could make a breakpoint and run it step by step. You will find the id (find it in "Variables" View) of "result" changed every time after java did
result = result + str;
On the other hand, if you use StringBuffer like
StringBuffer result = new StringBuffer("");
for(int i = 0; i < n; i++){
result.append(str);
}
the id of result will not change.

String objects are indeed immutable. result is not a String, it is a reference to a String object. In each iteration, a new String object is created and assigned to the same reference. The old object with no reference is eventually destroyed by a garbage collector. For a simple example like this, it is a possible solution. However, creating a new String object in each iteration in a real-world application is not a smart idea.

Using only 1 System.out.print() instead of 3. More details below [duplicate]

I am trying to concatenate strings in Java. Why isn't this working?
public class StackOverflowTest {
public static void main(String args[]) {
int theNumber = 42;
System.out.println("Your number is " . theNumber . "!");
}
}

You can concatenate Strings using the + operator:
System.out.println("Your number is " + theNumber + "!");
theNumber is implicitly converted to the String "42".

The concatenation operator in java is +, not .
Read this (including all subsections) before you start. Try to stop thinking the php way ;)
To broaden your view on using strings in Java - the + operator for strings is actually transformed (by the compiler) into something similar to:
new StringBuilder().append("firstString").append("secondString").toString()

There are two basic answers to this question:
[simple] Use the + operator (string concatenation). "your number is" + theNumber + "!" (as noted elsewhere)
[less simple]: Use StringBuilder (or StringBuffer).
StringBuilder value;
value.append("your number is");
value.append(theNumber);
value.append("!");
value.toString();
I recommend against stacking operations like this:
new StringBuilder().append("I").append("like to write").append("confusing code");
Edit: starting in java 5 the string concatenation operator is translated into StringBuilder calls by the compiler. Because of this, both methods above are equal.
Note: Spaceisavaluablecommodity,asthissentancedemonstrates.
Caveat: Example 1 below generates multiple StringBuilder instances and is less efficient than example 2 below
Example 1
String Blam = one + two;
Blam += three + four;
Blam += five + six;
Example 2
String Blam = one + two + three + four + five + six;

Out of the box you have 3 ways to inject the value of a variable into a String as you try to achieve:
1. The simplest way
You can simply use the operator + between a String and any object or primitive type, it will automatically concatenate the String and
In case of an object, the value of String.valueOf(obj) corresponding to the String "null" if obj is null otherwise the value of obj.toString().
In case of a primitive type, the equivalent of String.valueOf(<primitive-type>).
Example with a non null object:
Integer theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
Example with a null object:
Integer theNumber = null;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is null!
Example with a primitive type:
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
Output:
Your number is 42!
2. The explicit way and potentially the most efficient one
You can use StringBuilder (or StringBuffer the thread-safe outdated counterpart) to build your String using the append methods.
Example:
int theNumber = 42;
StringBuilder buffer = new StringBuilder()
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer.toString()); // or simply System.out.println(buffer)
Output:
Your number is 42!
Behind the scene, this is actually how recent java compilers convert all the String concatenations done with the operator +, the only difference with the previous way is that you have the full control.
Indeed, the compilers will use the default constructor so the default capacity (16) as they have no idea what would be the final length of the String to build, which means that if the final length is greater than 16, the capacity will be necessarily extended which has price in term of performances.
So if you know in advance that the size of your final String will be greater than 16, it will be much more efficient to use this approach to provide a better initial capacity. For instance, in our example we create a String whose length is greater than 16, so for better performances it should be rewritten as next:
Example optimized :
int theNumber = 42;
StringBuilder buffer = new StringBuilder(18)
.append("Your number is ").append(theNumber).append('!');
System.out.println(buffer)
Output:
Your number is 42!
3. The most readable way
You can use the methods String.format(locale, format, args) or String.format(format, args) that both rely on a Formatter to build your String. This allows you to specify the format of your final String by using place holders that will be replaced by the value of the arguments.
Example:
int theNumber = 42;
System.out.println(String.format("Your number is %d!", theNumber));
// Or if we need to print only we can use printf
System.out.printf("Your number is still %d with printf!%n", theNumber);
Output:
Your number is 42!
Your number is still 42 with printf!
The most interesting aspect with this approach is the fact that we have a clear idea of what will be the final String because it is much more easy to read so it is much more easy to maintain.

The java 8 way:
StringJoiner sj1 = new StringJoiner(", ");
String joined = sj1.add("one").add("two").toString();
// one, two
System.out.println(joined);
StringJoiner sj2 = new StringJoiner(", ","{", "}");
String joined2 = sj2.add("Jake").add("John").add("Carl").toString();
// {Jake, John, Carl}
System.out.println(joined2);

You must be a PHP programmer.
Use a + sign.
System.out.println("Your number is " + theNumber + "!");

"+" instead of "."

Use + for string concatenation.
"Your number is " + theNumber + "!"

This should work
public class StackOverflowTest
{
public static void main(String args[])
{
int theNumber = 42;
System.out.println("Your number is " + theNumber + "!");
}
}

For exact concatenation operation of two string please use:
file_names = file_names.concat(file_names1);
In your case use + instead of .

For better performance use str1.concat(str2) where str1 and str2 are string variables.

String.join( delimiter , stringA , stringB , … )
As of Java 8 and later, we can use String.join.
Caveat: You must pass all String or CharSequence objects. So your int variable 42 does not work directly. One alternative is using an object rather than primitive, and then calling toString.
Integer theNumber = 42;
String output =
String // `String` class in Java 8 and later gained the new `join` method.
.join( // Static method on the `String` class.
"" , // Delimiter.
"Your number is " , theNumber.toString() , "!" ) ; // A series of `String` or `CharSequence` objects that you want to join.
) // Returns a `String` object of all the objects joined together separated by the delimiter.
;
Dump to console.
System.out.println( output ) ;
See this code run live at IdeOne.com.

In java concatenate symbol is "+".
If you are trying to concatenate two or three strings while using jdbc then use this:
String u = t1.getString();
String v = t2.getString();
String w = t3.getString();
String X = u + "" + v + "" + w;
st.setString(1, X);
Here "" is used for space only.

In Java, the concatenation symbol is "+", not ".".

"+" not "."
But be careful with String concatenation. Here's a link introducing some thoughts from IBM DeveloperWorks.

You can concatenate Strings using the + operator:
String a="hello ";
String b="world.";
System.out.println(a+b);
Output:
hello world.
That's it

So from the able answer's you might have got the answer for why your snippet is not working. Now I'll add my suggestions on how to do it effectively. This article is a good place where the author speaks about different way to concatenate the string and also given the time comparison results between various results.
Different ways by which Strings could be concatenated in Java
By using + operator (20 + "")
By using concat method in String class
Using StringBuffer
By using StringBuilder
Method 1:
This is a non-recommended way of doing. Why? When you use it with integers and characters you should be explicitly very conscious of transforming the integer to toString() before appending the string or else it would treat the characters to ASCI int's and would perform addition on the top.
String temp = "" + 200 + 'B';
//This is translated internally into,
new StringBuilder().append( "" ).append( 200 ).append('B').toString();
Method 2:
This is the inner concat method's implementation
public String concat(String str) {
int olen = str.length();
if (olen == 0) {
return this;
}
if (coder() == str.coder()) {
byte[] val = this.value;
byte[] oval = str.value;
int len = val.length + oval.length;
byte[] buf = Arrays.copyOf(val, len);
System.arraycopy(oval, 0, buf, val.length, oval.length);
return new String(buf, coder);
}
int len = length();
byte[] buf = StringUTF16.newBytesFor(len + olen);
getBytes(buf, 0, UTF16);
str.getBytes(buf, len, UTF16);
return new String(buf, UTF16);
}
This creates a new buffer each time and copies the old content to the newly allocated buffer. So, this is would be too slow when you do it on more Strings.
Method 3:
This is thread safe and comparatively fast compared to (1) and (2). This uses StringBuilder internally and when it allocates new memory for the buffer (say it's current size is 10) it would increment it's 2*size + 2 (which is 22). So when the array becomes bigger and bigger this would really perform better as it need not allocate buffer size each and every time for every append call.
private int newCapacity(int minCapacity) {
// overflow-conscious code
int oldCapacity = value.length >> coder;
int newCapacity = (oldCapacity << 1) + 2;
if (newCapacity - minCapacity < 0) {
newCapacity = minCapacity;
}
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
return (newCapacity <= 0 || SAFE_BOUND - newCapacity < 0)
? hugeCapacity(minCapacity)
: newCapacity;
}
private int hugeCapacity(int minCapacity) {
int SAFE_BOUND = MAX_ARRAY_SIZE >> coder;
int UNSAFE_BOUND = Integer.MAX_VALUE >> coder;
if (UNSAFE_BOUND - minCapacity < 0) { // overflow
throw new OutOfMemoryError();
}
return (minCapacity > SAFE_BOUND)
? minCapacity : SAFE_BOUND;
}
Method 4
StringBuilder would be the fastest one for String concatenation since it's not thread safe. Unless you are very sure that your class which uses this is single ton I would highly recommend not to use this one.
In short, use StringBuffer until you are not sure that your code could be used by multiple threads. If you are damn sure, that your class is singleton then go ahead with StringBuilder for concatenation.

First method: You could use "+" sign for concatenating strings, but this always happens in print.
Another way: The String class includes a method for concatenating two strings: string1.concat(string2);

import com.google.common.base.Joiner;
String delimiter = "";
Joiner.on(delimiter).join(Lists.newArrayList("Your number is ", 47, "!"));
This may be overkill to answer the op's question, but it is good to know about for more complex join operations. This stackoverflow question ranks highly in general google searches in this area, so good to know.

you can use stringbuffer, stringbuilder, and as everyone before me mentioned, "+". I'm not sure how fast "+" is (I think it is the fastest for shorter strings), but for longer I think builder and buffer are about equal (builder is slightly faster because it's not synchronized).

here is an example to read and concatenate 2 string without using 3rd variable:
public class Demo {
public static void main(String args[]) throws Exception {
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(r);
System.out.println("enter your first string");
String str1 = br.readLine();
System.out.println("enter your second string");
String str2 = br.readLine();
System.out.println("concatenated string is:" + str1 + str2);
}
}

There are multiple ways to do so, but Oracle and IBM say that using +, is a bad practice, because essentially every time you concatenate String, you end up creating additional objects in memory. It will utilize extra space in JVM, and your program may be out of space, or slow down.
Using StringBuilder or StringBuffer is best way to go with it. Please look at Nicolas Fillato's comment above for example related to StringBuffer.
String first = "I eat"; String second = "all the rats.";
System.out.println(first+second);

Using "+" symbol u can concatenate strings.
String a="I";
String b="Love.";
String c="Java.";
System.out.println(a+b+c);

Separate Multiple Integers from a String

I've got a question about making a save function.
I'm trying to have a string be saved as a single file to set specific settings on a game. So saveFile would read "002007...", having 002 be a player's location, then 007 a player's level, for example.
I understand how to compile the various variables into a single string, but how would I return it to individual variables?

You better go with SQLite or SharedPreferences if you really want to save settings for a game on Android.
On the other hand, if you have to stick with saving a String on a file, you might want to use a delimiter(ie \r\n or # or | would do it) between numbers. So while parsing back delimiters will help you a lot, but beware when things get complicated a single String won't do the thing nicely. Then you might want to use JSON (for simplicity I would prefer gson) to encode your settings into one String and vice verse.

You could use a delimiter between the values like this:
int location = 02;
int level = 3;
int powerUps = 46;
... and so on
String saveString = location + "#" + level + "#" + powerUps + "#" + ...
Then to load the String back into variables:
String[] values = saveString.split("#");
location = values[0];
level = values[1];
powerUps = values[2];
... and so on

My advice is to check out Shared Preferences and you can read Android's documentation on it here.
If you did want to use your single String, file method, I suggest using delimiters. That simply means to put commas, or other types of delimeters in between different integer values. Instead of "002007", save it as "002,007". Example:
String s = "002,007"
String[] values = s.split(","); // values[0] is "002" and values[1] is "007"
Using the .split(String) command will return a String array with each element in the array containing parts of the String that was split up by the parameter, in this case: ,
If you wanted to separate values per person, something like this could be done:
String s = "002,007;003,008";
String[] people = s.split(";"); // people[0] is "002,007", people[1] is "003,004"
String[][] person = new String[people.length][people[0].split(",").length];
for (int i = 0; i < people.length; i++)
{
person[i] = people[i].split(",");
}
Here is what the array would then contain:
person[0][0] is "002"person[0][1] is "007" person[1][0] is "003" person[1][1] is "008"
// print it for your own testing
for (String ppl[] : person)
{
for (String val : ppl)
{
System.out.print(val + " ");
}
System.out.println("");
}

How do I concatenate input in java?

I am trying to concatenate and trying to parse at the same time. I am right now making a excel like program where I can say a1 = "Hello" + "World" and in the cell of A1 have it say HelloWorld. I just need to know how to parse the adding sign and connect those two words. Please tell me if you need more code to understand this, like the runner.
This is my parseInput class :
public class ParseInput {
private static String inputs;
static int col;
private static int row;
private static String operation;
private static Value field;
public static void parseInput(String input){
//splits the input at each regular expression match. \w is used for letters and \d && \D for integers
inputs = input;
Scanner tokens = new Scanner(inputs);
String none0 = tokens.next();
#SuppressWarnings("unused")
String none1 = tokens.next();
operation = tokens.nextLine().substring(1);
String[] holder = new String[2];
String regex = "(?<=[\\w&&\\D])(?=\\d)";
holder = none0.split(regex);
row = Integer.parseInt(holder[1]);
col = 0;
int counter = -1;
char temp = holder[0].charAt(0);
char check = 'a';
while(check <= temp){
if(check == temp){
col = counter +1;
}
counter++;
check = (char) (check + 1);
}
System.out.println(col);
System.out.println(row);
System.out.println(operation);
setField(Value.parseValue(operation));
Spreadsheet.changeCell(row, col, field);
}
public static Value getField() {
return field;
}
public static void setField(Value field) {
ParseInput.field = field;
}
}

This is actually a pretty complicated problem unless you can constrain input to a very small subset of what Excel accepts. If not then you'll probably want to look into something like ANTLR. However, assuming the above input then you'll want to do something like:
Split the string on the equal sign into s1 and s2
Split s2 on the plus sign into s3 and s4.
Trim all the strings, remove the quotes around s3 and s4.
Concatenate s3 and s4 and assign to your datastore indexed by s1.
Depending on how complex your concatenation needs are you can either use string concatenation or a StringBuilder:
result = "" + s3 + s4; // string concatenation
result = new StringBuilder().append(s3).append(s4).toString(); // StringBuilder
Let me know if you have any questions about any of the steps detailed above.
Details on (1) above, assuming input is a1 = "Hello" + "World":
String[] strings = input.split("=");
String s1 = strings[0].trim(); // a1
String s2 = strings[1].trim(); // "Hello" + "World"
strings = s2.split("+");
String s3 = strings[0].trim().replaceAll("^\"", "").replaceAll("\"$", "") // Hello
String s4 = strings[1].trim().replaceAll("^\"", "").replaceAll("\"$", ""); // World
String field = s3 + s4;
String colString = s1.replaceAll("[\\d]", ""); // a
String rowString = s1.replaceAll("[\\D]", ""); // 1
int col = colString.charAt(0) - 'a'; // 0
int row = Integer.parseInt(rowString);
Spreadsheet.changeCell(row, col, field);

I suggest you to implement your custom grammar using a parser generator like JavaCC.
Here you can find a simple tutorial.
I believe this is the better solution because in this way you can handle every expression you need.

Are you sure you want to use all the classes you are using? To parse something like "a=b+c+d.." (assuming you are not trying to validate), easiest and possibly the most efficient way is to use split API in Java lang String
Then join whatever is required using StringBuilder

You need to design and implement a parser and an evaluator. And before that, you need to design the language that your parser/evaluator is going to evaluate.
How to do it.
If your language is really simple, you can get away with parsing it by hand, using something like StringTokenizer to do the tokenization,
Otherwise, you are probably best off learning to use a Java "parser generator" such as JavaCC or ANTLR.
Either way, you need to do some background reading to understand all of the terminology. You could start with Wikipedia and/or the tutorial material from one of the parser generators. Alternatively, there are good textbooks on this topic.

In addition to what Abdullah said, if you really want to save every single ounce of memory you can, you should use the StringBuilder instead of the String concatenation. I believe i read somewhere before that the String concatenation make a new string object for each concatenations while the StringBuilder will add them all to a single String. Shouldn't matter too much though.

In my early life I made an equation evaluator in your style. It cost me huge code and complexity, because of my unawareness about Expression trees. But now with this you will be able to add more capabilities to your parser easily and with native JAVA codes. You will get tons of example of using Expression Trees.

Java - How to split (+ sign) on Strings so it could be read as a Variable/Int?

This question is rather difficult to confer, for simplistic sake:
I am loading some Strings via XML (XStream).
for example, Your total count is +variable+ .
The outcome would be
"Your total count is +variable+ ."
when it ideally should be
"Your total count is" + variable + "." aka "Your total count is 1."
The issue: (if you can't see it) it reads the variable as if it were a String.
I know I would need to split that String from where the plus sign starts and ends and then connect it to the String, for it to read as a variable, like the above. But how? I need this to be done so that the String before the variable and after it is split.
so:
"Your total count is 50, would you like a cookie?"
aka
"Your total count is " + variable + " , would you like a cookie?"
Thank you alot!
Okay, I agree it's very confusing. I've edited this post (read below).
Well I am loading some Strings via XML this could be the same case if I were loading them via a .txt or a config file.
On the XML file, I lay it out like so:
<list>
<dialogue>
<line>
<string> Your total count is + Somewhere.totalCount +, Would you like a cookie?</string>
</line>
</dialogue>
</list>
As you can see, the XML file can't locate where the variable (in a class is), nor can it recognise if it is a variable or a string.
I know that I would need to alter the way it reads it, so if there is a plus sign (+) anywhere on the String, it would simply "split" it away from the original String so I can reconnect it.
E.g. Your phone number is + PhoneBook.phoneNumber + should I call you? as it would be read from a XML file.
I want to "split" the String from front to back like so:
"Your phone number is " + PhoneBook.phoneNumber + " should I call you?"
At the same time, I'm not assigning a variable because It's already declared in the XML file, I want it to recognise it as a int.

First, Java can not know that the +variable+ part of your string should be replaced with the value of the corresponding variable and also does not provide some "eval" like functionality like PHP or other scripting languages do, which might help you with that.
If you want to exactly replace this specific '+variable+' part of the string, it can be done like this:
int variable = 1;
String text = "Your total count is +variable+.";
String textWithVariableValue = text.replaceAll("\\+variable\\+", Integer.toString(variable));
But if you want to replace variables with arbitrary names, you will have to put them into a Map first, and then find all occurences of +somename+ in the string and replace it with the corresponding value stored in the map. Something like this:
Map<String, Object> variables = new HashMap<String, Object>();
variables.put("var1", 1);
variables.put("foo", 5);
String text = "var1 = +var1+, foo = +foo+";
String textWithVariableValues = text;
for (String variableName : variables.keySet()) {
Object variableValue = variables.get(variableName);
textWithVariableValues = textWithVariableValues.replaceAll("\\+" + variableName + "\\+", variableValue.toString());
}

Sounds like what you need is the
String.format() method:
int total = calculateTotal();
String s = String.format("Your total is %1d.", total);

Not split, but find and replace.
Simplistically,
int variable = 1;
String src = "Your total count is +variable+.";
String result = src.replaceAll("\\+variable\\+.", Integer.toString(variable));
System.out.println(result);
Should print "Your total count is 1."
EDIT: (after your comment) If you need to replace a multiple variables in one go then the following works for me:
// Replace the ff. with the actual map of variables & values
Map<String, String> vars = Collections.singletonMap(
"variable", Integer.toString(123));
String src = "Your total count is +variable+.";
Pattern p = Pattern.compile("\\+(\\w+)\\+");
StringBuffer sb = new StringBuffer();
Matcher m = p.matcher(src);
while (m.find()) {
String varName = m.group(1);
if (vars.containsKey(varName)) {
m.appendReplacement(sb, vars.get(varName));
}
}
m.appendTail(sb);
System.out.println(sb.toString());
Prints "Your total count is 123."