当我使用代码报告错误时:
#include <stdio.h>
#include <string.h>
#include <regex.h>
int main(void)
{
//char * source = "change@/devices/soc/799999.i2c/i2c-3/3-0015/power_supply/battery";
char * regexString = "(?<action>[a-zA-Z]+)@\\/(?<dev_path>.*)\\/(?<subsystem>[a-zA-z]+)\\/(?<name>[a-zA-z]+)";
//size_t maxGroups = 4;
regex_t regexCompiled;
//regmatch_t groupArray[maxGroups];
int rc;
if ((rc = regcomp(®exCompiled, regexString, REG_EXTENDED)) != 0)
{
char buffer[1024];
regerror(rc, ®exCompiled, buffer, sizeof(buffer));
printf("Could not compile regular expression (%d: %s).\n", rc, buffer);
return 1;
}
regfree(®exCompiled);
return 0;
}
我得到了输出:
Could not compile regular expression (13: repetition-operator operand invalid).
问题在于您使用的符号(?:
"(?<action>[a-zA-Z]+)@\\/(?<dev_path>.*)\\/(?<subsystem>[a-zA-z]+)\\/(?<name>[a-zA-z]+)"
该符号适用于PCRE,而不是 POSIX。而 PCRE 在 ( 之后使用 ? 正是因为它在其他正则表达式系统(例如 POSIX)中无效。
所以,如果您想使用 PCRE 正则表达式,请安装并使用 PCRE 库。
否则,您需要使用:
"([a-zA-Z]+)@\\/(.*)\\/([a-zA-z]+)\\/([a-zA-z]+)"
有了这个,并注意到您需要一个 regmatch_t 来表示匹配的整个字符串以及 4 个捕获的组(总共 5 个捕获),您可以编写:
#include <stdio.h>
#include <string.h>
#include <regex.h>
int main(void)
{
char *source = "change@/devices/soc/799999.i2c/i2c-3/3-0015/power_supply/battery";
// char * regexString = "(?<action>[a-zA-Z]+)@\\/(?<dev_path>.*)\\/(?<subsystem>[a-zA-z]+)\\/(?<name>[a-zA-z]+)";
size_t maxGroups = 5;
char *regexString = "([a-zA-Z]+)@\\/(.*)\\/([a-zA-z]+)\\/([a-zA-z]+)";
regex_t regexCompiled;
regmatch_t groupArray[maxGroups];
int rc;
if ((rc = regcomp(®exCompiled, regexString, REG_EXTENDED)) != 0)
{
char buffer[1024];
regerror(rc, ®exCompiled, buffer, sizeof(buffer));
printf("Could not compile regular expression (%d: %s).\n", rc, buffer);
return 1;
}
if ((rc = regexec(®exCompiled, source, maxGroups, groupArray, 0)) != 0)
{
char buffer[1024];
regerror(rc, ®exCompiled, buffer, sizeof(buffer));
printf("Could not execute regular expression (%d: %s).\n", rc, buffer);
return 1;
}
printf("Match successful:\n");
for (size_t i = 0; i < maxGroups; i++)
{
int so = groupArray[i].rm_so;
int eo = groupArray[i].rm_eo;
printf("%zu: %d..%d [%.*s]\n", i, so, eo, eo - so, &source[so]);
}
regfree(®exCompiled);
return 0;
}
输出是:
Match successful:
0: 0..64 [change@/devices/soc/799999.i2c/i2c-3/3-0015/power_supply/battery]
1: 0..6 [change]
2: 8..43 [devices/soc/799999.i2c/i2c-3/3-0015]
3: 44..56 [power_supply]
4: 57..64 [battery]