Java - 创建一个具有指定长度并填充特定字符的新 String 实例。最佳解决方案？
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Java - Create a new String instance with specified length and filled with specific character. Best solution?
提问by Stefan Hendriks
I did check the other questions; this question has its focus on solving this particular question the most efficient way.
我确实检查了其他问题；这个问题的重点是以最有效的方式解决这个特定问题。
Sometimes you want to create a new string with a specified length, and with a default character filling the entire string.
有时您想创建一个具有指定长度的新字符串，并使用默认字符填充整个字符串。
ie, it would be cool if you could do new String(10, '*')and create a new String from there, with a length of 10 characters all having a *.
即，如果您可以new String(10, '*')从那里创建一个新的字符串，长度为 10 个字符的所有字符串都带有 * ，那就太酷了。
Because such a constructor does not exist, and you cannot extend from String, you have either to create a wrapper class or a method to do this for you.
因为这样的构造函数不存在，并且您不能从 String 扩展，所以您必须创建一个包装类或一个方法来为您执行此操作。
At this moment I am using this:
此时我正在使用这个：
protected String getStringWithLengthAndFilledWithCharacter(int length, char charToFill) {
char[] array = new char[length];
int pos = 0;
while (pos < length) {
array[pos] = charToFill;
pos++;
}
return new String(array);
}
It still lacks any checking (ie, when length is 0 it will not work). I am constructing the array first because I believe it is faster than using string concatination or using a StringBuffer to do so.
它仍然缺乏任何检查（即，当长度为 0 时，它将不起作用）。我首先构造数组，因为我相信它比使用字符串连接或使用 StringBuffer 这样做更快。
Anyone else has a better sollution?
其他人有更好的解决方案吗？
采纳答案by Cowan
Apache Commons Lang (probably useful enough to be on the classpath of any non-trivial project) has StringUtils.repeat():
Apache Commons Lang（可能足够有用以用于任何非平凡项目的类路径）具有StringUtils.repeat()：
String filled = StringUtils.repeat("*", 10);
Easy!
简单！
回答by Romain Linsolas
Simply use the StringUtils class from apache commons langproject. You have a leftPadmethod:
只需使用apache commons lang项目中的 StringUtils 类。你有一个leftPad方法：
StringUtils.leftPad("foobar", 10, '*'); // Returns "****foobar"
回答by unwind
No need to do the loop, and using just standard Java library classes:
无需执行循环，只需使用标准 Java 库类：
protected String getStringWithLengthAndFilledWithCharacter(int length, char charToFill) {
if (length > 0) {
char[] array = new char[length];
Arrays.fill(array, charToFill);
return new String(array);
}
return "";
}
As you can see, I also added suitable code for the length == 0case.
如您所见，我还为length == 0案例添加了合适的代码。
回答by Alexander Pavloshchuk
public static String fillString(int count,char c) {
StringBuilder sb = new StringBuilder( count );
for( int i=0; i<count; i++ ) {
sb.append( c );
}
return sb.toString();
}
What is wrong?
怎么了？
回答by Karl
To improve performance you could have a single predefined sting if you know the max length like:
为了提高性能，如果你知道最大长度，你可以有一个预定义的刺痛：
String template = "####################################";
字符串模板 = "##################################";
And then simply perform a substring once you know the length.
然后在知道长度后简单地执行一个子字符串。
回答by Michael Lloyd Lee mlk
The above is fine. Do you mind if I ask you a question - Is this causing you a problem? It seams to me you are optimizing before you know if you need to.
以上没问题。你介意我问你一个问题 - 这是否给你带来了问题？在我看来，您在知道是否需要之前正在优化。
Now for my over engineered solution. In many (thou not all) cases you can use CharSequence instead of a String.
现在是我过度设计的解决方案。在许多（不是全部）情况下，您可以使用 CharSequence 而不是 String。
public class OneCharSequence implements CharSequence {
private final char value;
private final int length;
public OneCharSequence(final char value, final int length) {
this.value = value;
this.length = length;
}
public char charAt(int index) {
if(index < length) return value;
throw new IndexOutOfBoundsException();
}
public int length() {
return length;
}
public CharSequence subSequence(int start, int end) {
return new OneCharSequence(value, (end-start));
}
public String toString() {
char[] array = new char[length];
Arrays.fill(array, value);
return new String(array);
}
}
回答by Chuim
One extra note: it seems that all public ways of creating a new Stringinstance involves necessarily the copy of whatever buffer you are working with, be it a char[], a StringBufferor a StringBuilder. From the Stringjavadoc (and is repeated in the respective toStringmethods from the other classes):
一个额外的注意事项：似乎所有创建新String实例的公共方法都必须涉及您正在使用的任何缓冲区的副本，无论是 a char[]、 aStringBuffer还是 a StringBuilder。来自Stringjavadoc（并toString在其他类的相应方法中重复）：
The contents of the character array are copied; subsequent modification of the character array does not affect the newly created string.
复制字符数组的内容；字符数组的后续修改不会影响新创建的字符串。
So you'll end up having a possibly big memory copy operation after the "fast filling" of the array. The only solution that may avoid this issue is the one from @mlk, if you can manage working directly with the proposed CharSequenceimplementation (what may be the case).
因此，在“快速填充”数组之后，您最终可能会进行大内存复制操作。唯一可以避免此问题的解决方案是来自 @mlk 的解决方案，前提是您可以直接使用建议的CharSequence实现进行管理（可能是这种情况）。
PS: I would post this as a comment but I don't have enough reputation to do that yet.
PS：我会将此作为评论发布，但我还没有足够的声誉来做到这一点。
回答by dfa
回答by Qadir Hussain
Try this Using the substring(int start, int end); method
试试这个 使用 substring(int start, int end); 方法
String myLongString = "abcdefghij";
if (myLongString .length() >= 10)
String shortStr = myLongString.substring(0, 5)+ "...";
this will return abcde.
这将返回 abcde。
回答by Pavel Netesa
char[] chars = new char[10];
Arrays.fill(chars, '*');
String text = new String(chars);
