Java 将单个字符转换为 CharSequence 的最有效方法
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Most efficient way to convert a single char to a CharSequence
提问by Graham Borland
What's the most efficient way to pass a single char to a method expecting a CharSequence?
将单个字符传递给需要 CharSequence 的方法的最有效方法是什么？
This is what I've got:
这就是我所拥有的：
textView.setText(new String(new char[] {c} ));
According to the answers given here, this is a sensible way of doing it where the input is a character array. I was wondering if there was a sneaky shortcut I could apply in the single-char case.
根据这里给出的答案，这是一种明智的做法，其中输入是字符数组。我想知道是否有可以在单字符情况下应用的偷偷摸摸的快捷方式。
采纳答案by eljenso
textView.setText(String.valueOf(c))
回答by Charles Goodwin
Shorthand, as in fewest typed characters possible:
速记，尽可能使用最少的输入字符：
c+""; // where c is a char
In full:
在全：
textView.setText(c+"");
回答by Grekz
char c = 'y';
textView.setText(""+c);
回答by trutheality
Looking at the implementation of the Character.toString(char c)method reveals that they use almost the same code you use:
查看该Character.toString(char c)方法的实现会发现它们使用的代码几乎与您使用的代码相同：
public String toString() {
char buf[] = {value};
return String.valueOf(buf);
}
For readability, you should just use Character.toString( c ).
为了可读性，您应该只使用Character.toString( c ).
Another efficient way would probably be
另一种有效的方法可能是
new StringBuilder(1).append(c);
It's definitely more efficient that using the +operator because, according to the javadoc:
+根据 javadoc 的说法，使用运算符肯定更有效：
The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings. String concatenation is implemented through the
StringBuilder(orStringBuffer) class and its append method
Java 语言为字符串连接运算符 ( + ) 以及将其他对象转换为字符串提供了特殊支持。字符串连接是通过
StringBuilder(orStringBuffer) 类及其 append 方法实现的
回答by phlogratos
A solution without concatenation is this:
没有串联的解决方案是这样的：
Character.valueOf(c).toString();
回答by Bruce Hamilton
The most compact CharSequence you can get when you have a handful of chars is the CharBuffer. To initialize this with your char value:
当您有少量字符时，您可以获得的最紧凑的 CharSequence 是 CharBuffer。要使用您的 char 值初始化它：
CharBuffer.wrap(new char[]{c});
That being said, using Strings is a fair bit more readable and easy to work with.
话虽如此，使用字符串更易读且易于使用。
