c - 释放 malloc 指针时出错

Question

在以下代码中释放 malloc 字符串时遇到错误:

long authenticate(char* user, long* uid, long* gid){
FILE* fp = fopen("/etc/passwd", "r");
char * line = NULL;
size_t len = 0;
long lineSize;
char* uname;

if (fp == NULL)
  exit(EXIT_FAILURE);

  while ((lineSize = getline(&line, &len, fp)) != -1) {

    uname = (char*)malloc(lineSize*sizeof(char));
    getInfo(line, lineSize, &uname, &uid, &gid);

    if(strcmp(uname, user) == 0) {  // Strings are equal
      printf("we've found a user!\n");
      printf("uname: %s uid: %ld gid: %ld\n", uname, *uid, *gid);
      free(uname);
      exit(1);
    }

    free(uname);
  }
}

每当我找到有效的用户名时，释放uname，然后退出，什么也没有发生。但是，当我在循环末尾尝试释放 uname 时找不到有效的 uname (if 语句未评估)时，我会得到以下信息错误:

malloc: *** error for object 0x10d20ef2c: pointer being freed was not allocated

我不明白为什么 free() 没有注册 uname 之前已进行 malloc。考虑到该错误的性质，我相信问题是我不了解malloc/free。

Answer 1

问题很可能是 getInfo()修改位置 uname指向，之后 free() 无法识别它.