我有一个txt格式的文档，假如是这样的：
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NULL, /* link_id = 5 */ \
NULL, /* link_id = 6 */ \
char*suit_case_am_num_01(void);
NULL, /* link_id = 12 */ \
NULL, /* link_id = 13 */ \
NULL, /* link_id = 14 */ \
NULL /* link_id = 15 */ \
char*suit_case_am_num_02(void);
char*suit_case_am_num_03(void);
NULL, /* link_id = 12 */ \
NULL, /* link_id = 13 */ \
NULL, /* link_id = 14 */ \
NULL /* link_id = 15 */ \
char*suit_case_am_num_04(void);
char*suit_case_am_num_05(void);
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我想将所有以char*开头，（void）结尾的中间那一段字符串截取出来：
最终希望得到的是这样的一串：
suit_case_am_num_01
suit_case_am_num_02
suit_case_am_num_03
suit_case_am_num_04
suit_case_am_num_05
改怎么写这段代码？
------解决方案--------------------------------------------------------
我就会笨办法，仅供参考。
- Java code
import java.util.regex.Matcher;import java.util.regex.Pattern;import java.io.*;public class FindStringWithChar{ public static void main(String[] args) { try { BufferedReader bf=new BufferedReader(new FileReader("3.txt")); String str=null; while((str=bf.readLine())!=null) { Pattern pattern = Pattern.compile("^char.*"); //找到char开头的字符串。 Matcher matcher = pattern.matcher(str); while (matcher.find()) { String s=matcher.group().replaceAll("char\\*","").replaceAll("\\(void\\);",""); //替换掉头"char*"和尾"(void);. System.out.println(s); } } } catch(Exception e) { e.printStackTrace(); } }}