如何通过C中的malloc为字符串分配指针数组？

Question

I have this struct in C Example:

typedef struct 
{
    const char * array_pointers_of_strings [ 30 ];
    // etc.
} message;

I need copy this array_pointers_of_strings to new array for sort strings. I need only copy adress.

while ( i < 30 )
{
   new_array [i] = new_message->array_pointers_of_strings [i]; 
   // I need only copy adress of strings
}

My question is: How to allocate new_array [i] by malloc() for only adress of strings?

Answer 1

As I can understand from your assignment statement in while loop I think you need array of strings instead:

char** new_array;
new_array = malloc(30 * sizeof(char*)); // ignore casting malloc

Note: By doing = in while loop as below:

new_array [i] = new_message->array_pointers_of_strings [i];

you are just assigning address of string (its not deep copy), but because you are also writing " only address of strings " so I think this is what you wants.

Edit: waring "assignment discards qualifiers from pointer target type"

you are getting this warning because you are assigning a const char* to char* that would violate the rules of const-correctness.

You should declare your new_array like:

const  char** new_array;      

or remove const in declaration of 'array_pointers_of_strings' from message stricture.

Answer 2

This:

char** p = malloc(30 * sizeof(char*));

will allocate a buffer big enough to hold 30 pointers to char (or string pointers, if you will) and assign to p its address.

p[0] is pointer 0, p[1] is pointer 1, ..., p[29] is pointer 29.

---

Old answer...

If I understand the question correctly, you can either create a fixed number of them by simply declaring variables of the type message :

message msg1, msg2, ...;

or you can allocate them dynamically:

message *pmsg1 = malloc(sizeof(message)), *pmsg2 = malloc(sizeof(message)), ...;

Answer 3

#include <stdio.h>
#include <stdlib.h>

#define ARRAY_LEN 2
typedef struct
{
    char * string_array [ ARRAY_LEN ];
} message;

int main() {
    int i;
    message message;
    message.string_array[0] = "hello";
    message.string_array[1] = "world";
    for (i=0; i < ARRAY_LEN; ++i ) {
        printf("%d %s\n",i, message.string_array[i]);
    }

    char ** new_message = (char **)malloc(sizeof(char*) * ARRAY_LEN);
    for (i=0; i < ARRAY_LEN; ++i ) {
        new_message[i] = message.string_array[i];
    }
    for (i=0; i < ARRAY_LEN; ++i ) {
        printf("%d %s\n",i, new_message[i]);
    }
}

Answer 4

is it mandatory for you to use Malloc? Because Calloc is the function in the C Standard Library which will make the job:

"The calloc() function allocates memory for an array of nmemb elements of size bytes each and returns a pointer to the allocated memory" . (Source: here )

I'm just creating a hash table and it has an array of pointers to nodes, and a easy way to do it is this:

hash_table_t *hash_table_create(unsigned long int size){
hash_table_t *ptr = NULL;

ptr = malloc(sizeof(hash_table_t) * 1);
if (ptr == NULL)
    return (NULL);

ptr->array = calloc(size, sizeof(hash_node_t *)); #HERE
if (ptr->array == NULL)
    return (NULL);
ptr->size = size;
return (ptr);}

Hope it works for you guys!

calloc c