Yes, it's a matter of style, because you'd expect sizeof(char) to always be one.

On the other hand, it's very much an idiom to use sizeof(foo) when doing a malloc, and most importantly it makes the code self documenting.

Also better for maintenance, perhaps. If you were switching from char to wchar, you'd switch to

Copywchar *p = malloc( sizeof(wchar) * ( len + 1 ) );

without much thought. Whereas converting the statement char *p = malloc( len + 1 ); would require more thought. It's all about reducing mental overhead.

And as @Nyan suggests in a comment, you could also do

Copytype *p = malloc( sizeof(*p) * ( len + 1 ) );

for zero-terminated strings and

Copytype *p = malloc( sizeof(*p) * len ) );

for ordinary buffers.

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.

As was indicated by others, you don't need to use malloc just to do:

const char *foo = "bar";

The reason for that is exactly that *foo is a pointer — when you initialize foo you're not creating a copy of the string, just a pointer to where "bar" lives in the data section of your executable. You can copy that pointer as often as you'd like, but remember, they're always pointing back to the same single instance of that string.

So when should you use malloc? Normally you use strdup() to copy a string, which handles the malloc in the background. e.g.

const char *foo = "bar";
char *bar = strdup(foo); /* now contains a new copy of "bar" */
printf("%s\n", bar);     /* prints "bar" */
free(bar);               /* frees memory created by strdup */

Now, we finally get around to a case where you may want to malloc if you're using sprintf() or, more safely snprintf() which creates / formats a new string.

char *foo = malloc(sizeof(char) * 1024);        /* buffer for 1024 chars */
snprintf(foo, 1024, "%s - %s\n", "foo", "bar"); /* puts "foo - bar\n" in foo */
printf(foo);                                    /* prints "foo - bar" */
free(foo);                                      /* frees mem from malloc */

char *array[10] declares an array of 10 pointers to char. It is not necessary to malloc storage for this array; it is embedded in struct List. Thus, the first call to malloc is unnecessary, as is the check immediately afterward.

The call to malloc inside the loop, and check after, are correct.

Also, in C, do not cast the return value of malloc; it can actually hide bugs.

Also also, sizeof(char) is 1 by definition and therefore you should never write it.

struct List
{
    char *array[10];
};

int
main(void)
{
    struct List input;
    for (int i = 0; i < 10; i++)
    {
        input.array[i] = malloc(10);
        if (!input.array[i])
            return EXIT_FAILURE;
    }
    /* presumably more code here */
    return 0;
}

Here are some things that may help you improve your code.

Fix the bug

Once memory is freed, it should not be referenced again. Unfortunately, your code allocates memory and then frees it and then returns a pointer to the freed memory. That's a serious bug! To fix it, simply omit the free within the function and make sure the caller calls free instead. Alternatively, you could avoid all of that by reversing the passed string in place.

Use the required #includes

The code uses strlen which means that it should #include <string.h> and malloc and free which means that it should #include <stdlib.h>. It was not difficult to infer, but it helps reviewers if the code is complete.

Use const where practical

In your revere_string routine, the string passed into the function is not and should not be altered. You should indicate that fact by declaring it like this:

char* reverse_string(const char* string)

Check for NULL pointers

The code must avoid dereferencing a NULL pointer if the call to malloc fails. The only indication that it has failed is if malloc returns NULL; if it does, it would probably make most sense to immediately return that NULL pointer.

Learn to use pointers instead of indexing

Using pointers effectively is an important C programming skill. This code could be made much simpler by doing an in-place reversal of the passed string and by using pointers:

char* reverse_string(char* string) {
    if (string == NULL) 
        return string;
    char *fwd = string;
    char *rev = &string[strlen(string)-1];

    while (rev > fwd) {
        char tmp = *rev;
        *rev-- = *fwd;
        *fwd++ = tmp;
    }
    return string;
}