如何将char转换为String?

时间:2011-11-17 18:37:19

标签: java string char type-conversion

我有char,我需要String。如何从一个转换为另一个?

12 个答案:

答案 0 :(得分:612)

您可以使用Character.toString(char)。请注意,此方法只会返回String.valueOf(char)的调用,这也有效。

正如其他人所说,字符串连接也可以作为捷径:

String s = "" + 's';

但这归结为:

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

效率较低,因为StringBuilderchar[]支持(由StringBuilder()过度分配到16),仅限于该数组被防御性地复制为结果String

String.valueOf(char)通过将char包装在单个元素数组中并将其传递给包私有构造函数String(char[], boolean)来“进入后门”,从而避免了数组副本。 / p>

答案 1 :(得分:220)

好问题。我有以下五种 6种方法。

1. String stringValueOf = String.valueOf('c'); // most efficient

2. String stringValueOfCharArray = String.valueOf(new char[]{x});

3. String characterToString = Character.toString('c');

4. String characterObjectToString = new Character('c').toString();

   // Although this method seems very simple, 
   // this is less efficient because the concatenation
   // expands to new StringBuilder().append(x).append("").toString();
5. String concatBlankString = 'c' + "";

6. String fromCharArray = new String(new char[]{x});
  

注意:Character.toString(char)会返回String.valueOf(char)。所以两者都是一样的。

String.valueOf(char[] value)调用new String(char[] value),后者又设置value char数组。

public String(char value[]) {
    this.value = Arrays.copyOf(value, value.length);
}

另一方面,String.valueOf(char value)调用以下包私有构造函数。

String(char[] value, boolean share) {
    // assert share : "unshared not supported";
    this.value = value;
}

来自Java 8 source code

String.java的源代码
  

因此,String.valueOf(char)似乎是在内存和速度方面最有效的方法,可以将char转换为String

  1. How to convert primitive char to String in Java
  2. How to convert Char to String in Java with Example

答案 2 :(得分:43)

以下是将char c转换为String s的各种方法(按速度和效率的降序排列)

char c = 'a';
String s = String.valueOf(c);             // fastest + memory efficient
String s = Character.toString(c);
String s = new String(new char[]{c});
String s = String.valueOf(new char[]{c});
String s = new Character(c).toString();
String s = "" + c;                        // slowest + memory inefficient

答案 3 :(得分:31)

使用以下任何一项:

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

答案 4 :(得分:24)

使用Character.toString()方法,如下所示:

char mChar = 'l';
String s = Character.toString(mChar);

答案 5 :(得分:15)

试试这个:Character.toString(aChar)或者只需:aChar + ""

答案 6 :(得分:12)

正如@WarFox所说 - 有6种方法可以将char转换为字符串。然而,最快的一个是通过连接,尽管上面的答案表明它是String.valueOf。以下是证明:

的基准
@BenchmarkMode(Mode.Throughput)
@Fork(1)
@State(Scope.Thread)
@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';

    @Benchmark
    public String stringValueOf() {
        return String.valueOf(c);
    }

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

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

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

    @Benchmark
    public String concatBlankStringPre() {
        return c + "";
    }

    @Benchmark
    public String concatBlankStringPost() {
        return "" + c;
    }

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

结果:

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

如您所见,最快的一个是c + """" + c;

VM version: JDK 1.8.0_131, VM 25.131-b11

此性能差异归因于-XX:+OptimizeStringConcat优化。你可以阅读它here

答案 7 :(得分:4)

我们有多种方法可以将public int getDecimal(String input) { int output = 0; // reverse input so that it is easier to work with input = new StringBuilder(input).reverse().toString(); for (int c = 0; c < input.length(); c++) { // check for invalid digit if (input.charAt(c) < '0' || input.charAt(c) > '7') return 0; if (input.charAt(c) != '0') { if (input.charAt(c) == '1') output += Math.pow(8, c); else // if it's greater than 1 output += Math.pow(8, c) + input.charAt(c) - 1; } } return output; } 转换为char一种方式是在String类中使用静态方法toString()

Character

实际上这个char ch = 'I'; String str1 = Character.toString(ch); 方法在内部使用来自toString类的valueOf方法,该方法使用了char数组:

String

所以第二方式是直接使用它:

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

String str2 = String.valueOf(ch); 类中的valueOf方法使用了char数组:

String

所以第三种方式是使用匿名数组来包装单个字符,然后将其传递给public static String valueOf(char c) { char data[] = {c}; return new String(data, true); } 构造函数:

String

第四种方式是使用连接:

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

这实际上将使用来自String str3 = "" + ch; 类的append方法,当我们在循环中进行连接时,它实际上是首选的。

答案 8 :(得分:4)

以下是一些方法,没有特别的顺序:

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
formatter.close();

答案 9 :(得分:2)

我正在将Char数组转换为String

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

答案 10 :(得分:0)

我已尝试过这些建议,但最终将其实施如下

editView.setFilters(new InputFilter[]{new InputFilter()
        {
            @Override
            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));
                    builder.append(source);
                    if (source instanceof Spanned)
                    {
                        TextUtils.copySpansFrom(
                                (Spanned) source, 0, source.length(), null, builder, newEnd);
                    }

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

答案 11 :(得分:0)

  char vIn = 'A';
  String vOut = Character.toString(vIn);

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