以下是一些方法，排名不分先后:

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();

你可以使用Character.toString(char)。注意，这个方法只是返回一个对String.valueOf(char)的调用，这也是有效的。

正如其他人所注意到的，字符串连接也可以作为一种快捷方式:

String s = "" + 's';

但这可以归结为:

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

这是较低的效率，因为StringBuilder是由char[]支持的(由StringBuilder()过度分配到16)，只有该数组才会被结果String防御性地复制。

String.valueOf(字符)通过将字符包装在单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”，这避免了数组复制。

试试这个:Character.toString(aChar)或这样:aChar + ""

我尝试了这些建议，但最终执行如下

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;
                }
            }
        }});

我有以下五六种方法。

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

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

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

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

// Method #5
// Although this approach seems very simple, 
// this is less efficient because the concatenation
// expands to a StringBuilder.
String concatBlankString = 'c' + "";

// Method #6
String fromCharArray = new String(new char[]{x});

注意:Character.toString(char)返回String.valueOf(char)。所以实际上两者是一样的。

字符串。valueOf(char[] value)调用new String(char[] value)，它反过来设置值char数组。

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

另一方面，字符串。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;
}

Java 8中的String.java源代码

因此，就内存和速度而言，String. valueof (char)似乎是将char转换为String的最有效的方法。

来源:

如何将原始字符转换为字符串在Java 如何转换字符到字符串在Java与示例

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

对于这些类型的转换，我有一个名为https://www.converttypes.com/的站点 它帮助我快速获得我使用的大多数语言的转换代码。