如何将char数组转换回字符串[重复]
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【中文标题】如何将char数组转换回字符串[重复]【英文标题】：how to convert a char array back into string [duplicate] 【发布时间】：2013-10-19 02:25:03 【问题描述】：我正在做一个搬运工词干分析器.....代码给了我 char 数组中的输出....但我需要将其转换为字符串以继续进行进一步的工作.....在我给出的代码中2 个单词“looking”和“walks”....返回为 look 和 walk（但在 char 数组中）...输出在 stem() 函数中打印
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package file;
import java.util.Vector;
/**
*
* @author sky
*/
public class stemmer
public static String line1;
private char[] b;
private int i, /* offset into b */
i_end, /* offset to end of stemmed word */
j, k;
private static final int INC = 50;
/* unit of size whereby b is increased */
public stemmer()
//b = new char[INC];
i = 0;
i_end = 0;
/**
* Add a character to the word being stemmed. When you are finished
* adding characters, you can call stem(void) to stem the word.
*/
public void add(char ch)
System.out.println("in add() function");
if (i == b.length)
char[] new_b = new char[i+INC];
for (int c = 0; c < i; c++)
new_b[c] = b[c];
b = new_b;
b[i++] = ch;
/** Adds wLen characters to the word being stemmed contained in a portion
* of a char[] array. This is like repeated calls of add(char ch), but
* faster.
*/
public void add(char[] w, int wLen)
if (i+wLen >= b.length)
char[] new_b = new char[i+wLen+INC];
for (int c = 0; c < i; c++)
new_b[c] = b[c];
b = new_b;
for (int c = 0; c < wLen; c++)
b[i++] = w[c];
public void addstring(String s1)
b=new char[s1.length()];
for(int k=0;k<s1.length();k++)
b[k] = s1.charAt(k);
System.out.println(b[k]);
i=s1.length();
/**
* After a word has been stemmed, it can be retrieved by toString(),
* or a reference to the internal buffer can be retrieved by getResultBuffer
* and getResultLength (which is generally more efficient.)
*/
public String toString() return new String(b,0,i_end);
/**
* Returns the length of the word resulting from the stemming process.
*/
public int getResultLength() return i_end;
/**
* Returns a reference to a character buffer containing the results of
* the stemming process. You also need to consult getResultLength()
* to determine the length of the result.
*/
public char[] getResultBuffer() return b;
/* cons(i) is true <=> b[i] is a consonant. */
private final boolean cons(int i)
switch (b[i])
case 'a': case 'e': case 'i': case 'o': case 'u': return false;
case 'y': return (i==0) ? true : !cons(i-1);
default: return true;
/* m() measures the number of consonant sequences between 0 and j. if c is
a consonant sequence and v a vowel sequence, and <..> indicates arbitrary
presence,
<c><v> gives 0
<c>vc<v> gives 1
<c>vcvc<v> gives 2
<c>vcvcvc<v> gives 3
....
*/
private final int m()
int n = 0;
int i = 0;
while(true)
if (i > j) return n;
if (! cons(i)) break; i++;
i++;
while(true)
while(true)
if (i > j) return n;
if (cons(i)) break;
i++;
i++;
n++;
while(true)
if (i > j) return n;
if (! cons(i)) break;
i++;
i++;
/* vowelinstem() is true <=> 0,...j contains a vowel */
private final boolean vowelinstem()
int i; for (i = 0; i <= j; i++) if (! cons(i)) return true;
return false;
/* doublec(j) is true <=> j,(j-1) contain a double consonant. */
private final boolean doublec(int j)
if (j < 1) return false;
if (b[j] != b[j-1]) return false;
return cons(j);
/* cvc(i) is true <=> i-2,i-1,i has the form consonant - vowel - consonant
and also if the second c is not w,x or y. this is used when trying to
restore an e at the end of a short word. e.g.
cav(e), lov(e), hop(e), crim(e), but
snow, box, tray.
*/
private final boolean cvc(int i)
if (i < 2 || !cons(i) || cons(i-1) || !cons(i-2)) return false;
int ch = b[i];
if (ch == 'w' || ch == 'x' || ch == 'y') return false;
return true;
private final boolean ends(String s)
int l = s.length();
int o = k-l+1;
if (o < 0)
return false;
for (int i = 0; i < l; i++)
if (b[o+i] != s.charAt(i))
return false;
j = k-l;
return true;
/* setto(s) sets (j+1),...k to the characters in the string s, readjusting
k. */
private final void setto(String s)
int l = s.length();
int o = j+1;
for (int i = 0; i < l; i++)
b[o+i] = s.charAt(i);
k = j+l;
/* r(s) is used further down. */
private final void r(String s) if (m() > 0) setto(s);
/* step1() gets rid of plurals and -ed or -ing. e.g.
caresses -> caress
ponies -> poni
ties -> ti
caress -> caress
cats -> cat
feed -> feed
agreed -> agree
disabled -> disable
matting -> mat
mating -> mate
meeting -> meet
milling -> mill
messing -> mess
meetings -> meet
*/
private final void step1()
if (b[k] == 's')
if (ends("sses")) k -= 2; else
if (ends("ies")) setto("i"); else
if (b[k-1] != 's') k--;
if (ends("eed")) if (m() > 0) k--; else
if ((ends("ed") || ends("ing")) && vowelinstem())
k = j;
if (ends("at")) setto("ate"); else
if (ends("bl")) setto("ble"); else
if (ends("iz")) setto("ize"); else
if (doublec(k))
k--;
int ch = b[k];
if (ch == 'l' || ch == 's' || ch == 'z') k++;
else if (m() == 1 && cvc(k)) setto("e");
/* step2() turns terminal y to i when there is another vowel in the stem. */
private final void step2() if (ends("y") && vowelinstem()) b[k] = 'i';
/* step3() maps double suffices to single ones. so -ization ( = -ize plus
-ation) maps to -ize etc. note that the string before the suffix must give
m() > 0. */
private final void step3() if (k == 0) return; /* For Bug 1 */ switch (b[k-1])
case 'a': if (ends("ational")) r("ate"); break;
if (ends("tional")) r("tion"); break;
break;
case 'c': if (ends("enci")) r("ence"); break;
if (ends("anci")) r("ance"); break;
break;
case 'e': if (ends("izer")) r("ize"); break;
break;
case 'l': if (ends("bli")) r("ble"); break;
if (ends("alli")) r("al"); break;
if (ends("entli")) r("ent"); break;
if (ends("eli")) r("e"); break;
if (ends("ousli")) r("ous"); break;
break;
case 'o': if (ends("ization")) r("ize"); break;
if (ends("ation")) r("ate"); break;
if (ends("ator")) r("ate"); break;
break;
case 's': if (ends("alism")) r("al"); break;
if (ends("iveness")) r("ive"); break;
if (ends("fulness")) r("ful"); break;
if (ends("ousness")) r("ous"); break;
break;
case 't': if (ends("aliti")) r("al"); break;
if (ends("iviti")) r("ive"); break;
if (ends("biliti")) r("ble"); break;
break;
case 'g': if (ends("logi")) r("log"); break;
/* step4() deals with -ic-, -full, -ness etc. similar strategy to step3. */
private final void step4() switch (b[k])
case 'e': if (ends("icate")) r("ic"); break;
if (ends("ative")) r(""); break;
if (ends("alize")) r("al"); break;
break;
case 'i': if (ends("iciti")) r("ic"); break;
break;
case 'l': if (ends("ical")) r("ic"); break;
if (ends("ful")) r(""); break;
break;
case 's': if (ends("ness")) r(""); break;
break;
/* step5() takes off -ant, -ence etc., in context <c>vcvc<v>. */
private final void step5()
if (k == 0) return; /* for Bug 1 */ switch (b[k-1])
case 'a': if (ends("al")) break; return;
case 'c': if (ends("ance")) break;
if (ends("ence")) break; return;
case 'e': if (ends("er")) break; return;
case 'i': if (ends("ic")) break; return;
case 'l': if (ends("able")) break;
if (ends("ible")) break; return;
case 'n': if (ends("ant")) break;
if (ends("ement")) break;
if (ends("ment")) break;
/* element etc. not stripped before the m */
if (ends("ent")) break; return;
case 'o': if (ends("ion") && j >= 0 && (b[j] == 's' || b[j] == 't')) break;
/* j >= 0 fixes Bug 2 */
if (ends("ou")) break; return;
/* takes care of -ous */
case 's': if (ends("ism")) break; return;
case 't': if (ends("ate")) break;
if (ends("iti")) break; return;
case 'u': if (ends("ous")) break; return;
case 'v': if (ends("ive")) break; return;
case 'z': if (ends("ize")) break; return;
default: return;
if (m() > 1) k = j;
/* step6() removes a final -e if m() > 1. */
private final void step6()
j = k;
if (b[k] == 'e')
int a = m();
if (a > 1 || a == 1 && !cvc(k-1)) k--;
if (b[k] == 'l' && doublec(k) && m() > 1) k--;
/** Stem the word placed into the Stemmer buffer through calls to add().
* Returns true if the stemming process resulted in a word different
* from the input. You can retrieve the result with
* getResultLength()/getResultBuffer() or toString().
*/
public void stem()
// step1();
// System.out.println("hello in stem");
// step2();
// step3();
// step4();
// step5();
// step6();
//
// i_end = k+1;
// i = 0;
System.out.println(i);
k = i - 1;
if (k > 1)
step1();
step2();
step3();
step4();
step5();
step6();
for(int c=0;c<=k;c++)
System.out.println(b[c]);
i_end = k+1; i = 0;
public static void main(String[] args)
stemmer s = new stemmer();
s.addstring("looking");
s.stem();
s.addstring("walks");
s.stem();
//System.out.println("Output " +s.b);
【问题讨论】：
【参考方案1】：char[] data = new char[10];
String text = String.valueOf(data);
【讨论】：
【参考方案2】：使用这种方式将 char[] 转换为字符串
String x=new String(char[])
例子
char x[]='a','m';
String z=new String(x);
System.out.println(z);
输出
am
【讨论】：
【参考方案3】： char[] a = new char[10];
for(int i=0;i<10;i++)
a[i] = 's';
System.out.println(new String(a));
或
System.out.println(String.copyValueOf(a));
【讨论】：
这里不需要与空字符串连接【参考方案4】：-使用Character类方法toString()；
例如：
class Test
public static void main (String[] args) throws java.lang.Exception
char c = 'a';
String s = Character.toString(c);
System.out.println(s);
- 现在使用上面解释的方法将所有字符数组项转换为字符串。
【讨论】：
问题是要求将char数组转换为字符串。不是字符到字符串 @newuser 亲爱的，您可以使用上述方法通过循环将 char 数组中的元素转换为 String .... @KumarVivekMitra 所以每次你连接字符串时以上是关于如何将char数组转换回字符串[重复]的主要内容，如果未能解决你的问题，请参考以下文章