Here is the shortest version (Java 1.5+ required):

repeated = new String(new char[n]).replace("\0", s); 

Where n is the number of times you want to repeat the string and s is the string to repeat.

No imports or libraries needed.

If you are using Java <= 7, this is as "concise" as it gets:

// create a string made up of n copies of string s String.format("%0" + n + "d", 0).replace("0", s); 

In Java 8 and above there is a more readable way:

// create a string made up of n copies of string s String.join("", Collections.nCopies(n, s)); 

Finally, for Java 11 and above, there is a new repeat​(int count) method specifically for this purpose(link)

"abc".repeat(12); 

Alternatively, if your project uses java libraries there are more options.

For Apache Commons:

StringUtils.repeat("abc", 12); 

For Google Guava:

Strings.repeat("abc", 12); 

Commons Lang StringUtils.repeat()

Usage:

String str = "abc"; String repeated = StringUtils.repeat(str, 3); repeated.equals("abcabcabc"); 

Java 8's String.join provides a tidy way to do this in conjunction with Collections.nCopies:

// say hello 100 times System.out.println(String.join("", Collections.nCopies(100, "hello"))); 

Here's a way to do it using only standard String functions and no explicit loops:

// create a string made up of n copies of s repeated = String.format(String.format("%%%ds", n), " ").replace(" ",s); 

If you're like me and want to use Google Guava and not Apache Commons. You can use the repeat method in the Guava Strings class.

Strings.repeat("-", 60); 

With java-8, you can also use Stream.generate.

import static java.util.stream.Collectors.joining; ... String repeated = Stream.generate(() -> "abc").limit(3).collect(joining()); //"abcabcabc" 

and you can wrap it in a simple utility method if needed:

public static String repeat(String str, int times) { return Stream.generate(() -> str).limit(times).collect(joining()); } 

So you want to avoid loops?

Here you have it:

public static String repeat(String s, int times) { if (times <= 0) return ""; else return s + repeat(s, times-1); } 

(of course I know this is ugly and inefficient, but it doesn't have loops :-p)

You want it simpler and prettier? use jython:

s * 3 

Edit: let's optimize it a little bit :-D

public static String repeat(String s, int times) { if (times <= 0) return ""; else if (times % 2 == 0) return repeat(s+s, times/2); else return s + repeat(s+s, times/2); } 

Edit2: I've done a quick and dirty benchmark for the 4 main alternatives, but I don't have time to run it several times to get the means and plot the times for several inputs... So here's the code if anybody wants to try it:

public class Repeat { public static void main(String[] args) { int n = Integer.parseInt(args[0]); String s = args[1]; int l = s.length(); long start, end; start = System.currentTimeMillis(); for (int i = 0; i < n; i++) { if(repeatLog2(s,i).length()!=i*l) throw new RuntimeException(); } end = System.currentTimeMillis(); System.out.println("RecLog2Concat: " + (end-start) + "ms"); start = System.currentTimeMillis(); for (int i = 0; i < n; i++) { if(repeatR(s,i).length()!=i*l) throw new RuntimeException(); } end = System.currentTimeMillis(); System.out.println("RecLinConcat: " + (end-start) + "ms"); start = System.currentTimeMillis(); for (int i = 0; i < n; i++) { if(repeatIc(s,i).length()!=i*l) throw new RuntimeException(); } end = System.currentTimeMillis(); System.out.println("IterConcat: " + (end-start) + "ms"); start = System.currentTimeMillis(); for (int i = 0; i < n; i++) { if(repeatSb(s,i).length()!=i*l) throw new RuntimeException(); } end = System.currentTimeMillis(); System.out.println("IterStrB: " + (end-start) + "ms"); } public static String repeatLog2(String s, int times) { if (times <= 0) { return ""; } else if (times % 2 == 0) { return repeatLog2(s+s, times/2); } else { return s + repeatLog2(s+s, times/2); } } public static String repeatR(String s, int times) { if (times <= 0) { return ""; } else { return s + repeatR(s, times-1); } } public static String repeatIc(String s, int times) { String tmp = ""; for (int i = 0; i < times; i++) { tmp += s; } return tmp; } public static String repeatSb(String s, int n) { final StringBuilder sb = new StringBuilder(); for(int i = 0; i < n; i++) { sb.append(s); } return sb.toString(); } } 

It takes 2 arguments, the first is the number of iterations (each function run with repeat times arg from 1..n) and the second is the string to repeat.

So far, a quick inspection of the times running with different inputs leaves the ranking something like this (better to worse):

1. Iterative StringBuilder append (1x).
2. Recursive concatenation log2 invocations (~3x).
3. Recursive concatenation linear invocations (~30x).
4. Iterative concatenation linear (~45x).

I wouldn't ever guessed that the recursive function was faster than the for loop :-o

Have fun(ctional xD).

This contains less characters than your question

public static String repeat(String s, int n) { if(s == null) { return null; } final StringBuilder sb = new StringBuilder(s.length() * n); for(int i = 0; i < n; i++) { sb.append(s); } return sb.toString(); } 

based on fortran's answer, this is a recusive version that uses a StringBuilder:

public static void repeat(StringBuilder stringBuilder, String s, int times) { if (times > 0) { repeat(stringBuilder.append(s), s, times - 1); } } public static String repeat(String s, int times) { StringBuilder stringBuilder = new StringBuilder(s.length() * times); repeat(stringBuilder, s, times); return stringBuilder.toString(); } 

using Dollar is simple as typing:

@Test public void repeatString() { String string = "abc"; assertThat($(string).repeat(3).toString(), is("abcabcabc")); } 

PS: repeat works also for array, List, Set, etc

I wanted a function to create a comma-delimited list of question marks for JDBC purposes, and found this post. So, I decided to take two variants and see which one performed better. After 1 million iterations, the garden-variety StringBuilder took 2 seconds (fun1), and the cryptic supposedly more optimal version (fun2) took 30 seconds. What's the point of being cryptic again?

private static String fun1(int size) { StringBuilder sb = new StringBuilder(size * 2); for (int i = 0; i < size; i++) { sb.append(",?"); } return sb.substring(1); } private static String fun2(int size) { return new String(new char[size]).replaceAll("\0", ",?").substring(1); } 

OOP Solution

Nearly every answer proposes a static function as a solution, but thinking Object-Oriented (for reusability-purposes and clarity) I came up with a Solution via Delegation through the CharSequence-Interface (which also opens up usability on mutable CharSequence-Classes).

The following Class can be used either with or without Separator-String/CharSequence and each call to "toString()" builds the final repeated String. The Input/Separator are not only limited to String-Class, but can be every Class which implements CharSequence (e.g. StringBuilder, StringBuffer, etc)!

Source-Code:

/** * Helper-Class for Repeating Strings and other CharSequence-Implementations * @author Maciej Schuttkowski */ public class RepeatingCharSequence implements CharSequence { final int count; CharSequence internalCharSeq = ""; CharSequence separator = ""; /** * CONSTRUCTOR - RepeatingCharSequence * @param input CharSequence to repeat * @param count Repeat-Count */ public RepeatingCharSequence(CharSequence input, int count) { if(count < 0) throw new IllegalArgumentException("Can not repeat String \""+input+"\" less than 0 times! count="+count); if(count > 0) internalCharSeq = input; this.count = count; } /** * CONSTRUCTOR - Strings.RepeatingCharSequence * @param input CharSequence to repeat * @param count Repeat-Count * @param separator Separator-Sequence to use */ public RepeatingCharSequence(CharSequence input, int count, CharSequence separator) { this(input, count); this.separator = separator; } @Override public CharSequence subSequence(int start, int end) { checkBounds(start); checkBounds(end); int subLen = end - start; if (subLen < 0) { throw new IndexOutOfBoundsException("Illegal subSequence-Length: "+subLen); } return (start == 0 && end == length()) ? this : toString().substring(start, subLen); } @Override public int length() { //We return the total length of our CharSequences with the separator 1 time less than amount of repeats: return count < 1 ? 0 : ( (internalCharSeq.length()*count) + (separator.length()*(count-1))); } @Override public char charAt(int index) { final int internalIndex = internalIndex(index); //Delegate to Separator-CharSequence or Input-CharSequence depending on internal index: if(internalIndex > internalCharSeq.length()-1) { return separator.charAt(internalIndex-internalCharSeq.length()); } return internalCharSeq.charAt(internalIndex); } @Override public String toString() { return count < 1 ? "" : new StringBuilder(this).toString(); } private void checkBounds(int index) { if(index < 0 || index >= length()) throw new IndexOutOfBoundsException("Index out of Bounds: "+index); } private int internalIndex(int index) { // We need to add 1 Separator-Length to total length before dividing, // as we subtracted one Separator-Length in "length()" return index % ((length()+separator.length())/count); } } 

Usage-Example:

public static void main(String[] args) { //String input = "12345"; //StringBuffer input = new StringBuffer("12345"); StringBuilder input = new StringBuilder("123"); //String separator = "<=>"; StringBuilder separator = new StringBuilder("<=");//.append('>'); int repeatCount = 2; CharSequence repSeq = new RepeatingCharSequence(input, repeatCount, separator); String repStr = repSeq.toString(); System.out.println("Repeat="+repeatCount+"\tSeparator="+separator+"\tInput="+input+"\tLength="+input.length()); System.out.println("CharSeq:\tLength="+repSeq.length()+"\tVal="+repSeq); System.out.println("String :\tLength="+repStr.length()+"\tVal="+repStr); //Here comes the Magic with a StringBuilder as Input, as you can append to the String-Builder //and at the same Time your Repeating-Sequence's toString()-Method returns the updated String :) input.append("ff"); System.out.println(repSeq); //Same can be done with the Separator: separator.append("===").append('>'); System.out.println(repSeq); } 

Example-Output:

Repeat=2 Separator=<= Input=123 Length=3 CharSeq: Length=8 Val=123<=123 String : Length=8 Val=123<=123 123ff<=123ff 123ff<====>123ff 

using only JRE classes (System.arraycopy) and trying to minimize the number of temp objects you can write something like:

public static String repeat(String toRepeat, int times) { if (toRepeat == null) { toRepeat = ""; } if (times < 0) { times = 0; } final int length = toRepeat.length(); final int total = length * times; final char[] src = toRepeat.toCharArray(); char[] dst = new char[total]; for (int i = 0; i < total; i += length) { System.arraycopy(src, 0, dst, i, length); } return String.copyValueOf(dst); } 

EDIT

and without loops you can try with:

public static String repeat2(String toRepeat, int times) { if (toRepeat == null) { toRepeat = ""; } if (times < 0) { times = 0; } String[] copies = new String[times]; Arrays.fill(copies, toRepeat); return Arrays.toString(copies). replace("[", ""). replace("]", ""). replaceAll(", ", ""); } 

EDIT 2

using Collections is even shorter:

public static String repeat3(String toRepeat, int times) { return Collections.nCopies(times, toRepeat). toString(). replace("[", ""). replace("]", ""). replaceAll(", ", ""); } 

however I still like the first version.

Not the shortest, but (i think) the fastest way is to use the StringBuilder:

 /** * Repeat a String as many times you need. * * @param i - Number of Repeating the String. * @param s - The String wich you want repeated. * @return The string n - times. */ public static String repeate(int i, String s) { StringBuilder sb = new StringBuilder(); for (int j = 0; j < i; j++) sb.append(s); return sb.toString(); } 

If speed is your concern, then you should use as less memory copying as possible. Thus it is required to work with arrays of chars.

public static String repeatString(String what, int howmany) { char[] pattern = what.toCharArray(); char[] res = new char[howmany * pattern.length]; int length = pattern.length; for (int i = 0; i < howmany; i++) System.arraycopy(pattern, 0, res, i * length, length); return new String(res); } 

To test speed, a similar optimal method using StirngBuilder is like this:

public static String repeatStringSB(String what, int howmany) { StringBuilder out = new StringBuilder(what.length() * howmany); for (int i = 0; i < howmany; i++) out.append(what); return out.toString(); } 

and the code to test it:

public static void main(String... args) { String res; long time; for (int j = 0; j < 1000; j++) { res = repeatString("123", 100000); res = repeatStringSB("123", 100000); } time = System.nanoTime(); res = repeatString("123", 1000000); time = System.nanoTime() - time; System.out.println("elapsed repeatString: " + time); time = System.nanoTime(); res = repeatStringSB("123", 1000000); time = System.nanoTime() - time; System.out.println("elapsed repeatStringSB: " + time); } 

And here the run results from my system:

elapsed repeatString: 6006571 elapsed repeatStringSB: 9064937 

Note that the test for loop is to kick in JIT and have optimal results.

a straightforward one-line solution:
requires Java 8

Collections.nCopies( 3, "abc" ).stream().collect( Collectors.joining() ); 

for the sake of readability and portability:

public String repeat(String str, int count){ if(count <= 0) {return "";} return new String(new char[count]).replace("\0", str); } 

Seeing a number of StringBuilder related answers, I believe it's also worth mentioning that there are new StringBuilder.repeat and StringBuffer.repeat methods added in Java 21 to simplify the appending of multiple copies of characters or strings:

System.out.println(new StringBuilder().repeat('A', 10)); // AAAAAAAAAA System.out.println(new StringBuilder().repeat("A1", 2)); // A1A1 

If you are worried about performance, just use a StringBuilder inside the loop and do a .toString() on exit of the loop. Heck, write your own Util class and reuse it. It will have five lines of code at most.

I really enjoy this question. There is a lot of knowledge and styles. So I can't leave it without show my rock and roll ;)

{ String string = repeat("1234567890", 4); System.out.println(string); System.out.println("======="); repeatWithoutCopySample(string, 100000); System.out.println(string);// This take time, try it without printing System.out.println(string.length()); } /** * The core of the task. */ @SuppressWarnings("AssignmentToMethodParameter") public static char[] repeat(char[] sample, int times) { char[] r = new char[sample.length * times]; while (--times > -1) { System.arraycopy(sample, 0, r, times * sample.length, sample.length); } return r; } /** * Java classic style. */ public static String repeat(String sample, int times) { return new String(repeat(sample.toCharArray(), times)); } /** * Java extreme memory style. */ @SuppressWarnings("UseSpecificCatch") public static void repeatWithoutCopySample(String sample, int times) { try { Field valueStringField = String.class.getDeclaredField("value"); valueStringField.setAccessible(true); valueStringField.set(sample, repeat((char[]) valueStringField.get(sample), times)); } catch (Exception ex) { throw new RuntimeException(ex); } } 

Do you like it?

public static String repeat(String str, int times) { int length = str.length(); int size = length * times; char[] c = new char[size]; for (int i = 0; i < size; i++) { c[i] = str.charAt(i % length); } return new String(c); } 

Simple loop

public static String repeat(String string, int times) { StringBuilder out = new StringBuilder(); while (times-- > 0) { out.append(string); } return out.toString(); } 

Try this out:

public static char[] myABCs = {'a', 'b', 'c'}; public static int numInput; static Scanner in = new Scanner(System.in); public static void main(String[] args) { System.out.print("Enter Number of Times to repeat: "); numInput = in.nextInt(); repeatArray(numInput); } public static int repeatArray(int y) { for (int a = 0; a < y; a++) { for (int b = 0; b < myABCs.length; b++) { System.out.print(myABCs[b]); } System.out.print(" "); } return y; } 

Using recursion, you can do the following (using ternary operators, one line max):

public static final String repeat(String string, long number) { return number == 1 ? string : (number % 2 == 0 ? repeat(string + string, number / 2) : string + repeat(string + string, (number - 1) / 2)); } 

I know, it's ugly and probably not efficient, but it's one line!

If you only know the length of the output string (and it may be not divisible by the length of the input string), then use this method:

static String repeat(String s, int length) { return s.length() >= length ? s.substring(0, length) : repeat(s + s, length); } 

Usage demo:

for (int i = 0; i < 50; i++) System.out.println(repeat("_/‾\\", i)); 

Don't use with empty s and length > 0, since it's impossible to get the desired result in this case.

Despite your desire not to use loops, I think you should use a loop.

String repeatString(String s, int repetitions) { if(repetitions < 0) throw SomeException(); else if(s == null) return null; StringBuilder stringBuilder = new StringBuilder(s.length() * repetitions); for(int i = 0; i < repetitions; i++) stringBuilder.append(s); return stringBuilder.toString(); } 

Your reasons for not using a for loop are not good ones. In response to your criticisms:

1. Whatever solution you use will almost certainly be longer than this. Using a pre-built function only tucks it under more covers.
2. Someone reading your code will have to figure out what you're doing in that non-for-loop. Given that a for-loop is the idiomatic way to do this, it would be much easier to figure out if you did it with a for loop.
3. Yes someone might add something clever, but by avoiding a for loop you are doing something clever. That's like shooting yourself in the foot intentionally to avoid shooting yourself in the foot by accident.
4. Off-by-one errors are also mind-numbingly easy to catch with a single test. Given that you should be testing your code, an off-by-one error should be easy to fix and catch. And it's worth noting: the code above does not contain an off-by-one error. For loops are equally easy to get right.
5. So don't reuse variables. That's not the for-loop's fault.
6. Again, so does whatever solution you use. And as I noted before; a bug hunter will probably be expecting you to do this with a for loop, so they'll have an easier time finding it if you use a for loop.

Consolidated for quick reference:

public class StringRepeat { // Java 11 has built-in method - str.repeat(3); // Apache - StringUtils.repeat(3); // Google - Strings.repeat("",n); // System.arraycopy static String repeat_StringBuilderAppend(String str, int n) { if (str == null || str.isEmpty()) return str; StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { sb.append(str); } return sb.toString(); } static String repeat_ArraysFill(String str, int n) { String[] strs = new String[n]; Arrays.fill(strs, str); return Arrays.toString(strs).replaceAll("\\[|\\]|,| ", ""); } static String repeat_Recursion(String str, int n) { if (n <= 0) return ""; else return str + repeat_Recursion(str, n - 1); } static String repeat_format1(String str, int n) { return String.format(String.format("%%%ds", n), " ").replace(" ", str); } static String repeat_format2(String str, int n) { return new String(new char[n]).replace("\0", str); } static String repeat_format3(String str, int n) { return String.format("%0" + n + "d", 0).replace("0", str); } static String repeat_join(String str, int n) { return String.join("", Collections.nCopies(n, str)); } static String repeat_stream(String str, int n) { return Stream.generate(() -> str).limit(n).collect(Collectors.joining()); } public static void main(String[] args) { System.out.println(repeat_StringBuilderAppend("Mani", 3)); System.out.println(repeat_ArraysFill("Mani", 3)); System.out.println(repeat_Recursion("Mani", 3)); System.out.println(repeat_format1("Mani", 3)); System.out.println(repeat_format2("Mani", 3)); System.out.println(repeat_format3("Mani", 3)); System.out.println(repeat_join("Mani", 3)); System.out.println(repeat_stream("Mani", 3)); } 

}

here is the latest Stringutils.java StringUtils.java

 public static String repeat(String str, int repeat) { // Performance tuned for 2.0 (JDK1.4) if (str == null) { return null; } if (repeat <= 0) { return EMPTY; } int inputLength = str.length(); if (repeat == 1 || inputLength == 0) { return str; } if (inputLength == 1 && repeat <= PAD_LIMIT) { return repeat(str.charAt(0), repeat); } int outputLength = inputLength * repeat; switch (inputLength) { case 1 : return repeat(str.charAt(0), repeat); case 2 : char ch0 = str.charAt(0); char ch1 = str.charAt(1); char[] output2 = new char[outputLength]; for (int i = repeat * 2 - 2; i >= 0; i--, i--) { output2[i] = ch0; output2[i + 1] = ch1; } return new String(output2); default : StringBuilder buf = new StringBuilder(outputLength); for (int i = 0; i < repeat; i++) { buf.append(str); } return buf.toString(); } } 

it doesn't even need to be this big, can be made into this, and can be copied and pasted into a utility class in your project.

 public static String repeat(String str, int num) { int len = num * str.length(); StringBuilder sb = new StringBuilder(len); for (int i = 0; i < times; i++) { sb.append(str); } return sb.toString(); } 

So e5, I think the best way to do this would be to simply use the above mentioned code,or any of the answers here. but commons lang is just too big if it's a small project