posix正则表达式，多个匹配结果，为何只取到了第一个？解决办法

posix正则表达式，多个匹配结果，为何只取到了第一个？？？
如题，明明有多个匹配的结果，为何运行结果是只匹配了第一个呢？还望高人指点!

用的是Fedora12系统中自带的regex.h文件。

源码及运行结果：

C/C++ code

#include <stdio.h>#include <string.h>#include <sys/types.h>#include <regex.h>// 提取子串char* getsubstr(char *s, regmatch_t *pmatch){    static char buf[100] = {0};    memset(buf, 0, sizeof(buf));    memcpy(buf, s+pmatch->rm_so, pmatch->rm_eo - pmatch->rm_so);    return buf;}int main(int argc, char **argv){    int status, i;    int cflags = REG_EXTENDED;    regmatch_t pmatch[5];    const size_t nmatch = 5;    regex_t reg;    const char *pattern = "[[:lower:]]+";        // 正则表达式    char buf[] = "COMEdavID2012@gmail.com";        // 待搜索的字符串    regcomp(&reg, pattern, cflags);    status = regexec(&reg, buf, nmatch, pmatch, 0);    if(status == REG_NOMATCH)        printf("No Match\n");    else    {        printf("Match:\n");        for(i = 0; i < nmatch; i++)        {            if(pmatch[i].rm_so == -1)                continue;            char *p = getsubstr(buf, &pmatch[i]);            printf("[%d, %d): %s\n", pmatch[i].rm_so, pmatch[i].rm_eo, p);        }    }    regfree(&reg);    return 0;}

运行结果：
[zcm@t #139]$make
gcc -c -o a.o a.c
gcc -o a a.o
[zcm@t #140]$./a
Match:
[4, 7): dav
[zcm@t #141]$



[解决办法]
不清楚啊。
perl里是 /[a-z]+/g
[解决办法]
regexec() is used to match a null-terminated string against the precompiled pattern buffer, preg. nmatch and　pmatch are used to provide information regarding the location of any matches. eflags may be the bitwise-or　of one or both of REG_NOTBOL and REG_NOTEOL which cause changes in matching behavior described below.

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