带有字符指针的 strsep() 在 c 中给出分段错误

Question

int main() {
  char* instruction = "x10,   14(x2)";
  char* rd = strsep(&instruction, ",");
  printf("%c", *rd);
    
  return(0);
}

我试图运行这几行，我希望它打印“x”，但它给出了分段错误。 我做错了什么？

Answer 1

从strsep 人

 char *strsep(char **stringp, const char *delim); If *stringp is NULL, the strsep() function returns NULL and does nothing else. Otherwise, this function finds the first token in the string *stringp, that is delimited by one of the bytes in the string delim. This token is terminated by overwriting the delimiter with a null byte ('\\0'), and *stringp is updated to point past the token. In case no delimiter was found, the token is taken to be the entire string *stringp, and *stringp is made NULL.

您正试图通过将其传递给strsep来修改仅就绪的字符串。

  char* instruction = "x10,   14(x2)";
  char* rd = strsep(&instruction, ",");

尝试

  char* instruction = "x10,   14(x2)";
  char *temp = strdup(instruction);
  char *save = temp;
  char* rd = strsep(&temp, ",");
  free(save);

您需要保留指向temp的指针的副本，以便以后可以将其释放。

示例：来自描述“ stringp 已更新为指向令牌之后。” 如果您使用free(temp) ，你会经过一个无效的地址， free()因为temp不再指向分配块的某处点在它的内部的开始。