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Java String Concatenation with + operator
(5 answers)
Closed 3 years ago.
Consider the below code snippet -
public class Student {
public static void main(String args[]) {
int a = 3;
int b = 4;
System.out.println(a + b +" ");
System.out.println(a + b +" "+ a+b);
System.out.println(""+ a+b);
}
}
The output for the above snippet is coming as -
7
7 34
34
It is clear from the output that if we use String at first in the print statement then the integers are concatenated. But, if we use integer at first then the values are added and displayed.
Can someone please explain why is this behavior?
I even tried to look at the implementation of println() method in PrintStream class but could not figure out.
Actually it's not println() implementation who's causing that, this is Java way to treat the + operator when dealing with Strings.
In fact the operation is treated from left to right so:
If the string comes before int (or any other type) in the operation String conversion is used and all the rest of the operands will be treated as strings, and it consists only of a String concatenation operation.
If int comes first in the operation it will be treated as int, thus addition operation is used.
That's why a + b +" " gives 7 because Stringis in the end of the operation, and for other expressions a + b +" "+ a+b or ""+ a+b, the variables a and b will be treated as strings if the come after a String in the expression.
For further details you can check String Concatenation Operator + in Java Specs.
It has nothing to do with the implementation of println. The value of the expression passed to println is evaluated before println is executed.
It is evaluated left to right. As long as both operands of + are numeric, addition will be performed. Once at least one of the operands is a String, String concatenation will be performed.
System.out.println(a + b + " ");
// int + int int + String
// addition, concatenation
System.out.println(a + b + " " + a + b);
// int + int int + String Str+Str Str+Str
// addition, concat, concat, concat
System.out.println("" + a + b);
// String+int String+int
// concat, concat
First case:
System.out.println(a + b +" ");
In this order, a + b will first be evaluated as a int sum, then converted to a string when adding the space (the second + here is for string concatenation).
Second case:
System.out.println(a + b +" "+ a+b);
Here the first part will be int sum operation then converted to a string as we add the space (the 2nd + is for string concatenation), the rest will be string concatenation as the left operand is already a string.
Third case:
System.out.println(""+ a+b);
Same as 2nd.
Notes:
In order to change this behavior, just add parenthesis to force the int sum before the string concatenations.
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);
This question already has answers here:
Concatenating null strings in Java [duplicate]
(5 answers)
Closed 7 years ago.
How to add whichever character between every character of a premade String? (JAVA)
For example, I have the String "Hello world" and I have to add '_' between every character of the String.
Any function or useful code I can use to do it?
I have to do an algorithm that make me output "H_e_l_l_o_ _w_o_r_l_d"
This is what I have:
public String example(String s) {
String s2 = null;
for(int i = 0; i < s.length(); i++){
s2 += s.charAt(i) + (((i+1) == 0) ? " " : "-");
}
return s2;
}
My output in the main class is being:
nullH-e-l-l-o- -w-o-r-l-d-
Don't know why
This looks like a homework assignment. So, I won't directly write out all the code.
String = "hello world";
Say, there is a variable len = str.length() - 1. Instead of doing it from index 0, we will start our for loop from len - 1. The character 'd' is at index len, and the '_' will have to be inserted right before that. This can be done by setting the string to str = str.substring(0,i) + "_" + str.substring(i+1);
You will have to use a for loop that starts from len - 1 and goes on till the index reached is 0.
Now, on every single iteration, when you are inserting a character assigning str to str.substring(0,i) + "_" + str.substring(i+1); causes you to make a new string object, which is absolutely horrible style. This can be solved by using a StringBuilder.
Does that make it clear?
In the future, refrain from posting questions without having done any work. This community is there to help you with solving issues that you may have in your solutions, not write your solutions for you.
I'm slightly confused on this test question. I made a chart of the values of i , j, and the string. I got "nbearig", but my runtime is printing out numbers. I'm not sure where I went wrong. ++i , --j means that they were incre/decremented before the code after the for loop right?
public class AlGore {
public static void main(String[] args) {
String mystery = "mnerigpaba";
String solved = "";
int len = mystery.length();
for (int i = 0, j = len - 1; i < len/2; ++i, --j) {
solved += mystery.charAt(i) + mystery.charAt(j);
}
System.out.println(solved);
}
}
I'm not sure where I went wrong. ++i , --j means that they were incre/decremented before the code after the for loop right?
1) They were preincremented / predecremented respectively.
2) It happened after each execution of the loop body.
my compiler is printing out numbers.
No it isn't. The compiler is compiling your code!!! The JVM is printing numbers ... when you run the code.
To understand the reason why, take a careful look at this:
solved += mystery.charAt(i) + mystery.charAt(j);
This is equivalent to
solved = solved + ( mystery.charAt(i) + mystery.charAt(j) );
Now the expression in brackets performs a numeric addition of a character to a character. According to the rules of Java expressions, that gives an int value. So the entire expression becomes:
solved = String.concat(
solved,
Integer.toString(mystery.charAt(i) + mystery.charAt(j));
I thought that the charAt(i) function will return a string?
No. It returns a char ... just like the method name "charAt" implies. String and char are fundamentally different types.
Comment: That is a good exam question, it tests how well you understand loops, and how well you understand Java expression semantics.
You are performing integer math (because char is an integral type),
// solved += mystery.charAt(i) + mystery.charAt(j);
solved += Character.toString(mystery.charAt(i))
+ Character.toString(mystery.charAt(j));
That way you are performing String concatenation.
mystery.charAt(i) + mystery.charAt(j); will add the numeric values of those two characters. You can force string concatenation by adding "" + before:
solved += "" + mystery.charAt(i) + mystery.charAt(j);
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Can I multiply strings in java to repeat sequences?
In Python, we can easily multiply the stings.
count = 10
print '*' * count
Is there any similar option available in Java?
You can use Dollar for your purposes(Java API that unifies collections, arrays, iterators/iterable, and char sequences.)
String str = $("*").repeat(count);
In this way you will get result "**********" as you want.
Java doesn't have that feature for now.
char[10] c = new char[10];
Arrays.fill(c, '*');
String str = new String(c);
To avoid creating a new String everytime.
How about this??
System.out.println(String.format("%10s", "").replace(' ', '*'));
This gives me output as **********.
I believe this is what you want...
Update 1
int yournumber = 10;
System.out.println(String.format("%" + yournumber + "s","*").replace(' ', '*'));
Good Luck!!!
The simplest way to do that in Java is to use a for() loop:
String s = "";
for (int i = 0; i < 10; i++) {
s += "*";
}
System.out.println(s);