Allocated Array

With an allocated array it's straightforward enough to follow.

Declare your array of pointers. Each element in this array points to a struct Test:

struct Test *array[50];

Then allocate and assign the pointers to the structures however you want. Using a loop would be simple:

array[n] = malloc(sizeof(struct Test));

Then declare a pointer to this array:

                               // an explicit pointer to an array 
struct Test *(*p)[] = &array;  // of pointers to structs

This allows you to use (*p)[n]->data; to reference the nth member.

Don't worry if this stuff is confusing. It's probably the most difficult aspect of C.

Dynamic Linear Array

If you just want to allocate a block of structs (effectively an array of structs, not pointers to structs), and have a pointer to the block, you can do it more easily:

struct Test *p = malloc(100 * sizeof(struct Test));  // allocates 100 linear
                                                     // structs

You can then point to this pointer:

struct Test **pp = &p

You don't have an array of pointers to structs any more, but it simplifies the whole thing considerably.

Dynamic Array of Dynamically Allocated Structs

The most flexible, but not often needed. It's very similar to the first example, but requires an extra allocation. I've written a complete program to demonstrate this that should compile fine.

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

struct Test {
    int data;
};

int main(int argc, char **argv)
{
    srand(time(NULL));

    // allocate 100 pointers, effectively an array
    struct Test **t_array = malloc(100 * sizeof(struct Test *));

    // allocate 100 structs and have the array point to them
    for (int i = 0; i < 100; i++) {
        t_array[i] = malloc(sizeof(struct Test));
    }

    // lets fill each Test.data with a random number!
    for (int i = 0; i < 100; i++) {
        t_array[i]->data = rand() % 100;
    }

    // now define a pointer to the array
    struct Test ***p = &t_array;
    printf("p points to an array of pointers.\n"
       "The third element of the array points to a structure,\n"
       "and the data member of that structure is: %d\n", (*p)[2]->data);

    return 0;
}

Output:

> p points to an array of pointers.
> The third element of the array points to a structure,
> and the data member of that structure is: 49

Or the whole set:

for (int i = 0; i < 100; i++) {
    if (i % 10 == 0)
        printf("\n");
    printf("%3d ", (*p)[i]->data);
}

 35  66  40  24  32  27  39  64  65  26 
 32  30  72  84  85  95  14  25  11  40 
 30  16  47  21  80  57  25  34  47  19 
 56  82  38  96   6  22  76  97  87  93 
 75  19  24  47  55   9  43  69  86   6 
 61  17  23   8  38  55  65  16  90  12 
 87  46  46  25  42   4  48  70  53  35 
 64  29   6  40  76  13   1  71  82  88 
 78  44  57  53   4  47   8  70  63  98 
 34  51  44  33  28  39  37  76   9  91 

Dynamic Pointer Array of Single-Dynamic Allocated Structs

This last example is rather specific. It is a dynamic array of pointers as we've seen in previous examples, but unlike those, the elements are all allocated in a single allocation. This has its uses, most notable for sorting data in different configurations while leaving the original allocation undisturbed.

We start by allocating a single block of elements as we do in the most basic single-block allocation:

struct Test *arr = malloc(N*sizeof(*arr));

Now we allocate a separate block of pointers:

struct Test **ptrs = malloc(N*sizeof(*ptrs));

We then populate each slot in our pointer list with the address of one of our original array. Since pointer arithmetic allows us to move from element to element address, this is straight-forward:

for (int i=0;i<N;++i)
    ptrs[i] = arr+i;

At this point the following both refer to the same element field

arr[1].data = 1;
ptrs[1]->data = 1;

And after review the above, I hope it is clear why.

When we're done with the pointer array and the original block array they are freed as:

free(ptrs);
free(arr);

Note: we do NOT free each item in the ptrs[] array individually. That is not how they were allocated. They were allocated as a single block (pointed to by arr), and that is how they should be freed.

So why would someone want to do this? Several reasons.

First, it radically reduces the number of memory allocation calls. Rather then N+1 (one for the pointer array, N for individual structures) you now have only two: one for the array block, and one for the pointer array. Memory allocations are one of the most expensive operations a program can request, and where possible, it is desirable to minimize them (note: file IO is another, fyi).

Another reason: Multiple representations of the same base array of data. Suppose you wanted to sort the data both ascending and descending, and have both sorted representations available at the same time. You could duplicate the data array, but that would require a lot of copying and eat significant memory usage. Instead, just allocate an extra pointer array and fill it with addresses from the base array, then sort that pointer array. This has especially significant benefits when the data being sorted is large (perhaps kilobytes, or even larger, per item) The original items remain in their original locations in the base array, but now you have a very efficient mechanism in which you can sort them without having to actually move them. You sort the array of pointers to items; the items don't get moved at all.

I realize this is an awful lot to take in, but pointer usage is critical to understanding the many powerful things you can do with the C language, so hit the books and keep refreshing your memory. It will come back.

The syntax you are looking for is somewhat cumbersome, but it looks like this:

// Declare test_array_ptr as pointer to array of test_t
test_t (*test_array_ptr)[];

You can then use it like so:

test_array_ptr = &array_t1;
(*test_array_ptr)[0] = new_struct;

To make the syntax easier to understand, you can use a typedef:

// Declare test_array as typedef of "array of test_t"
typedef test_t test_array[];
...
// Declare test_array_ptr as pointer to test_array
test_array *test_array_ptr = &array_t1;
(*test_array_ptr)[0] = new_struct;

The cdecl utility is useful for deciphering complex C declarations, especially when arrays and function pointers get involved.

Remember that unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" (T [N]) will be converted ("decay") to an expression of type "pointer to T" (T *), and the value of the expression will be the address of the first element of the array.

This naturally leads us to the following:

tTest *ptr = sTestStructure;  // note no & operator!

The expression sTestStructure is implicitly converted from type "20-element array of tTest" to "pointer to tTest", and the value of the expression is the same as &sTestStructure[0]. To access any element of sTestStructure, you could simply index into ptr as you would sTestStructure:

ptr[i].u8Mem1 = some_value;                // sTestStructure[i].u8Mem1 = some_value
printf( "%" PRIu8 "\n", ptr[j].u8Mem2 );

Remember that the subscript operation a[i] is defined as *(a + i); given the address a, offset i elements from that address and defereference the result. Thus, the [] operator implicitly dereferences ptr, which is why we use the . operator to access each struct member.

Alternately, you could also access struct members using the -> operator, and advance the pointer as necessary:

tTest *ptr = sTestStructure;
while( ptr->u8Mem1 != some_value )  // no subscript operation here
  ptr++;

The expression ptr->u8Mem1 is equivalent to (*ptr).u8Mem1¹, which is equivalent to ptr[0].u8Mem1.

So what happens if we decide to use &sTestStructure instead? Since sTestStructure is the operand of the unary & operator, the conversion rule above doesn't apply; instead of getting a pointer to a pointer to tTest, we get a pointer to a 20-element array of tTest, or:

tTest (*arrPtr)[20] = &sTestStructure;

This presents a bit more of a challenge, since we have to dereference arrPtr before we can index into it:

(*arrPtr)[i].u8Mem1 = some_value;
printf( "%" PRIu8 "\n", (*arrPtr)[j].u8Mem2 );

Since a[i] is defined as *(a + i), the expression *arrPtr can also be written as arrPtr[0] (*arrPtr == *(arrPtr + 0) == arrPtr[0]). So those lines could also be written as

arrPtr[0][i].u8Mem1 = some_value;
printf( "%" PRIu8 "\n", arrPtr[0][j].u8Mem2 );

As should be evident from those last couple of lines, you normally don't see this form of an array pointer unless you're dealing with multi-dimensional arrays. Remember our conversion rule, where an expression of type T [N] is converted to an expression of type T *? Well, replace T with an array type like Q[M], and we get the conversion from "N-element array of M-element array of Q" (Q [N][M]) to "pointer to M-element array of Q" (Q (*)[M]):

tTest sTestStructure[10][20];
tTest (*arrPtr)[20] = sTestStructure;

Looks good. Just assign it to something of the same type, struct Country. Also, as pointed out in the comments, it should be malloc num_of_countries * sizeof struct Country (not the pointer type), which is now correctly dereferenced below as sizeof (*countries_array) which also works.

pCountry countries_array;
countries_array = malloc(num_of_countries * sizeof (*countries_array));
struct Country Jefferson = {"Jefferson", 1,2,3 };
countries_array[0] = Jefferson;

// don't forget to free the memory when no longer needed.
free (countries_array);

If we must put a pointer into this array of structs, we can either dereference the pointer like countries_array[0] = *pointer, or... we could declare countries_array as an array of pointers, instead of an array of structs. Perhaps this is what you may want. Either way, the actual structures have to occupy memory somewhere...

pCountry *countries_array = malloc(num_of_countries*sizeof countries_array);
pCountry j = &Jefferson; // `&`, "address of" operator
countries_array[0] = j; // put a `pointer` into the array...

"Is it true, that structure Array contains array of structures Object?" No it isn't. Structure Array contains a pointer, not an array. To create an arry you have to allocate memory with malloc or calloc and then assign it to this pointer.

"If I use alloc in function, how can I free it later, if I don't know, how many times I have called that function containing malloc?" There is free( void* ptr ) to clean the memory you've allocated with malloc or calloc. Have a look here for example - http://www.cplusplus.com/reference/cstdlib/free/

If you want to have an array inside you must have something like this

typedef struct {
  unsigned size;
  Object items[10];
} Array;

It will have junk in the memory. To check if pointer is NULL do: if (pointer==NULL) or more idiomatically if (!pointer).

Since you want arrays, you need to declare arrays:

char *book[] = { "x", "y", "z",};
int number[] = { 1, 2, 3};

Another issue is

list = (struct data*) malloc( sizeof(struct data) );

//assigning arguments
list[count]->bookname = ...

Here, list is always going to have exactly one element. So if count is anything other than 0, you will be accessing an array out of bounds!