How to allocate array of pointers for strings by malloc in C?

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.

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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)), ...;