Analysis on the function of pointer

This article mainly introduces a usage of pointers to pointers. Therefore, we will not talk about pointers or pointer pointing to them here. Their basic code is directly used (Purpose: Use a function to dynamically apply for memory, and assign values. Output the value after calling the function.) [cpp] # include <stdio. h> # include <stdlib. h> typedef struct Data {int da;} Data; void fun (Data * p); int main () {Data * d; fun (d ); printf ("% d", d-> da); return 0;} void fun (Data * p) {p = (Data *) malloc (sizeof (Data )); p-> da = 2;} I suggest you run it directly to check the results, and you will find an error. Here I suggest you think about it first and then look at the following code: [cpp] # include <stdio. h> # include <stdlib. h> typedef struct Data {int da;} Data; void fun (Data ** p); // here Int main () {Data * d; fun (& d); // printf ("% d", d-> da); return 0 ;} void fun (Data ** p) // here {(* p) = (Data *) malloc (sizeof (Data); // here (* p) -> da = 2; // changed here} It is not difficult to find out that the changed Code uses a pointer pointing to the pointer. Why can't I use the pointer directly? Www.2cto.com because the void fun (Data * p) parameter in the first code is passed by value. The call function automatically defines a temporary variable to store the copy of the real parameter. The memory dynamically allocated in fun is not pointed to by the pointer variable d in the main function, the pointer variable d in main is undefined from start to end. How can this problem be solved? Use a pointer to the pointer to pass in the address of the pointer variable d in main. In this way, the memory space address allocated in fun is stored in the pointer variable d. Here, one of the functions of the pointer is self-evident.