Question

我有一个功能：

func (struct passwd* pw)
{

struct passwd* temp;
struct passwd* save;

temp = getpwnam("someuser");
/* since getpwnam returns a pointer to a static 
 * data buffer, I am copying the returned struct
 * to a local struct.
 */

if(temp) {
   save = malloc(sizeof *save);
   if (save) {
       memcpy(save, temp, sizeof(struct passwd));
       /* Here, I have to update passed pw* with this save struct. */
       *pw = *save; /* (~ memcpy) */
   }
}

}

调用func（pw）的函数能够获取更新的信息。

但如上所述可以使用它。声明* pw = * save不是深层副本。我不想像逐个复制每个结构的每个成员 pw-＆gt; pw_shell = strdup（save-＆gt; pw_shell）等。

有没有更好的方法呢？

感谢。

Answer 1

函数参数必须为struct passwd**，然后更改*passwd

Answer 2

如果你愿意，你可以做一个浅拷贝，但结果只有在下次调用getpenam之前才会有效。但为什么要复制两次？你的malloc是内存泄漏！这样做会很好：

void func (struct passwd *pw)
{
  struct passwd *tmp = getpenam("someuser"); // get a pointer to a static struct
  *pw = *tmp;  // copy the struct to caller's storage.
}

如果您需要深层复制，则必须逐字段进行：

void deep_func (struct passwd *pw)
{
  struct passwd *tmp = getpenam("someuser"); // get a pointer to a static struct
  *pw = *tmp; // copy everything
  pw->pw_name = safe_strdup(pw->pw_name);  // Copy pointer contents.
  pw->pw_passwd = safe_strdup(pw->pw_passwd);
  // etc for all pointer fields
}

对于深层复制，您需要一个相应的例程来释放malloc（）'ed storage：

void free_passwd_fields(struct passwd *pw)
{
  free(pw->pw_name);
  free(pw->pw_passwd);
  // etc
}

进行通话的好方法是：

// Declare a 1-element array of structs.  
// No &'s are needed, so code is simplified, and a later change to malloc()/free() is very simple.
struct passwd pw[1];

// ... and later
func(pw);

// pw now behaves like a pointer to a struct, but with no malloc or free needed.
// For example:
printf("login name is %s\n", pw->pw_name);

// Done with copy.  Free it.
free_passwd_fields(pw);

如何在指向它的指针时填充结构作为参数传递给函数

2 个答案: