如何为双指针分配内存，并且它应该在f1，f2和f3中可见？

Question

My sample program is below.

void function1 () {
    void **data
    function2(data);
    //Will access data here
}

void function2(void **data) {
    function3(data);
    //Will access data here
}

void function3(void **data) {
    //Need to allocate memory here says 100 bytes for data
    //Populate some values in data
}

My actual need:

1. void* should be allocated in function1
2. that should be passed across function2 and function3
3. memory must be allocated in function3 only.
4. data must be accessed in function2 and function3

Could you please help me how to do this?

Thanks, Boobesh

Answer 1

OP expresses conflicting needs for data

In function1() , data is a " pointer to pointer to void".
Yet in function3() OP wants data to be "... 100 bytes for data ".

A more typical paradigm is that data in function1() is

void function1 () {
  void *data = NULL;
  function2(&data);  // Address of data passed, so funciton2() receives a void **
  //Will access data here
  unsigned char *ucp = (unsigned char *) data;
  if ((ucp != NULL) && (ucp[0] == 123)) { 
    ;  // success
  }
  ...
  // Then when done
  free(data);
  data = 0;
}

Then in this case the memory allocation for data in function3() is

void function3(void **data) {
  if (data == NULL) {
    return;
  }
  // Need to allocate memory here.  Say 100 bytes for data
  size_t Length = 100;
  *data = malloc(Length);
  if (data != NULL) {
    //Populate some values in data
    memset(*data, 123, Length);
  }
}

void function2(void **data) {
  if (data == NULL) {
    return;
  }
  function3(data);
  unsigned char *ucp = (unsigned char *) *data;
  // Will access data here
  if ((ucp != NULL) &&  (ucp[0] == 123)) { 
    ;  // success
  }
}