java - array - How to convert a char to a String?

how to store character in string in java (8)

I have a char and I need a String. How do I convert from one to the other?

I am converting Char Array to String

Char[] CharArray={ 'A', 'B', 'C'};
String text = String.copyValueOf(CharArray);

As @WarFox stated - there are 6 methods to convert char to string. However, the fastest one would be via concatenation, despite answers above stating that it is String.valueOf. Here is benchmark that proves that:

@Warmup(iterations = 10, time = 1, batchSize = 1000, timeUnit = TimeUnit.SECONDS)
@Measurement(iterations = 10, time = 1, batchSize = 1000, timeUnit = TimeUnit.SECONDS)
public class CharToStringConversion {

    private char c = 'c';

    public String stringValueOf() {
        return String.valueOf(c);

    public String stringValueOfCharArray() {
        return String.valueOf(new char[]{c});

    public String characterToString() {
        return Character.toString(c);

    public String characterObjectToString() {
        return new Character(c).toString();

    public String concatBlankStringPre() {
        return c + "";

    public String concatBlankStringPost() {
        return "" + c;

    public String fromCharArray() {
        return new String(new char[]{c});

And result:

Benchmark                                        Mode  Cnt       Score      Error  Units
CharToStringConversion.characterObjectToString  thrpt   10   82132.021 ± 6841.497  ops/s
CharToStringConversion.characterToString        thrpt   10  118232.069 ± 8242.847  ops/s
CharToStringConversion.concatBlankStringPost    thrpt   10  136960.733 ± 9779.938  ops/s
CharToStringConversion.concatBlankStringPre     thrpt   10  137244.446 ± 9113.373  ops/s
CharToStringConversion.fromCharArray            thrpt   10   85464.842 ± 3127.211  ops/s
CharToStringConversion.stringValueOf            thrpt   10  119281.976 ± 7053.832  ops/s
CharToStringConversion.stringValueOfCharArray   thrpt   10   86563.837 ± 6436.527  ops/s

As you can see, the fastest one would be c + "" or "" + c;

VM version: JDK 1.8.0_131, VM 25.131-b11

This performance difference is due to -XX:+OptimizeStringConcat optimization. You can read about it here.

Here are a few methods, in no particular order:

char c = 'c';

String s = Character.toString(c); // Most efficient way

s = new Character(c).toString(); // Same as above except new Character objects needs to be garbage-collected

s = c + ""; // Least efficient and most memory-inefficient, but common amongst beginners because of its simplicity

s = String.valueOf(c); // Also quite common

s = String.format("%c", c); // Not common

Formatter formatter = new Formatter();
s = formatter.format("%c", c).toString(); // Same as above

I've tried the suggestions but ended up implementing it as follows

editView.setFilters(new InputFilter[]{new InputFilter()
            public CharSequence filter(CharSequence source, int start, int end,
                                       Spanned dest, int dstart, int dend)
                String prefix = "http://";

                //make sure our prefix is visible
                String destination = dest.toString();

                //Check If we already have our prefix - make sure it doesn't
                //get deleted
                if (destination.startsWith(prefix) && (dstart <= prefix.length() - 1))
                    //Yep - our prefix gets modified - try preventing it.
                    int newEnd = (dend >= prefix.length()) ? dend : prefix.length();

                    SpannableStringBuilder builder = new SpannableStringBuilder(
                            destination.substring(dstart, newEnd));
                    if (source instanceof Spanned)
                                (Spanned) source, 0, source.length(), null, builder, newEnd);

                    return builder;
                    //Accept original replacement (by returning null)
                    return null;

Try this: Character.toString(aChar) or just this: aChar + ""

Use any of the following:

String str = String.valueOf('c');
String str = Character.toString('c');
String str = 'c' + "";

We have various ways to convert a char to String. One way is to make use of static method toString() in Character class:

char ch = 'I'; 
String str1 = Character.toString(ch);

Actually this toString method internally makes use of valueOf method from String class which makes use of char array:

public static String toString(char c) {
    return String.valueOf(c);

So second way is to use this directly:

String str2 = String.valueOf(ch);

This valueOf method in String class makes use of char array:

public static String valueOf(char c) {
        char data[] = {c};
        return new String(data, true);

So the third way is to make use of an anonymous array to wrap a single character and then passing it to String constructor:

String str4 = new String(new char[]{ch});

The fourth way is to make use of concatenation:

String str3 = "" + ch;

This will actually make use of append method from StringBuilder class which is actually preferred when we are doing concatenation in a loop.

You can use Character.toString(char). Note that this method simply returns a call to String.valueOf(char), which also works.

As others have noted, string concatenation works as a shortcut as well:

String s = "" + 's';

But this compiles down to:

String s = new StringBuilder().append("").append('s').toString();

which is less efficient because the StringBuilder is backed by a char[] (over-allocated by StringBuilder() to 16), only for that array to be defensively copied by the resulting String.

String.valueOf(char) "gets in the back door" by wrapping the char in a single-element array and passing it to the package private constructor String(char[], boolean), which avoids the array copy.