My favorite:

#include <stdlib.h>

struct st *x = malloc(sizeof *x); 

Note that:

• x must be a pointer
• no cast is required
• include appropriate header

No, you're not allocating memory for y->x twice.

Instead, you're allocating memory for the structure (which includes a pointer) plus something for that pointer to point to.

Think of it this way:

         1          2
        +-----+    +------+
y------>|  x------>|  *x  |
        |  n  |    +------+
        +-----+

You actually need the two allocations (1 and 2) to store everything you need.

Additionally, your type should be struct Vector *y since it's a pointer, and you should never cast the return value from malloc in C.

It can hide certain problems you don't want hidden, and C is perfectly capable of implicitly converting the void* return value to any other pointer.

And, of course, you probably want to encapsulate the creation of these vectors to make management of them easier, such as with having the following in a header file vector.h:

#include <stddef.h>

struct Vector {
    double *data;    // Use readable names rather than x/n.
    size_t size;
};

struct Vector *newVector(size_t sz);
void delVector(struct Vector *vector);
//void setVectorItem(struct Vector *vector, size_t idx, double val);
//double getVectorItem(struct Vector *vector, size_t idx);

Then, in vector.c, you have the actual functions for managing the vectors:

#include <stdlib.h>
#include "vector.h"

// Atomically allocate a two-layer object. Either both layers
// are allocated or neither is, simplifying memory checking.

struct Vector *newVector(size_t sz) {
    // First, the vector layer.

    struct Vector *vector = malloc(sizeof (struct Vector));
    if (vector == NULL)
        return NULL;

    // Then data layer, freeing vector layer if fail.

    vector->data = malloc(sz * sizeof (double));
    if (vector->data == NULL) {
        free(vector);
        return NULL;
    }

    // Here, both layers worked. Set size and return.

    vector->size = sz;
    return vector;
}

void delVector(struct Vector *vector) {
    // Can safely assume vector is NULL or fully built.

    if (vector != NULL) {
        free(vector->data);
        free(vector);
    }
}

By encapsulating the vector management like that, you ensure that vectors are either fully built or not built at all - there's no chance of them being half-built.

It also allows you to totally change the underlying data structures in future without affecting clients. For example:

• if you wanted to make them sparse arrays to trade off space for speed.
• if you wanted the data saved to persistent storage whenever changed.
• if you wished to ensure all vector elements were initialised to zero.
• if you wanted to separate the vector size from the vector capacity for efficiency⁽¹⁾.

You could also add more functionality such as safely setting or getting vector values (see commented code in the header), as the need arises.

For example, you could (as one option) silently ignore setting values outside the valid range and return zero if getting those values. Or you could raise an error of some description, or attempt to automatically expand the vector under the covers⁽¹⁾.

In terms of using the vectors, a simple example is something like the following (very basic) main.c

#include "vector.h"

int main(void) {
    struct Vector *myvec = newVector(42);
    myvec->data[0] = 2.718281828459;
    delVector(myvec);
}

⁽¹⁾ That potential for an expandable vector bears further explanation.

Many vector implementations separate capacity from size. The former is how many elements you can use before a re-allocation is needed, the latter is the actual vector size (always <= the capacity).

When expanding, you want to generally expand in such a way that you're not doing it a lot, since it can be an expensive operation. For example, you could add 5% more than was strictly necessary so that, in a loop continuously adding one element, it doesn't have to re-allocate for every single item.

If you're just using a simple struct, you don't need more memory allocated for it as time goes on. You just create the struct, use it, and clean it up if required. If you are dynamically allocating your struct (ie: with malloc), then you test the value of the pointer-to-struct you create and see if it is NULL. If it is NULL, then the memory allocation failed, and you can either retry, or abandon further operations (ie: exit on error condition).

#include <stdio.h>

typedef struct myStruct {
  int i;
  char c;
} myStruct;

int main(void) {
  // Static allocation, no cleanup required
  myStruct staticStruct;
  staticStruct.i = 0;
  staticStruct.c = 'c';

  // Dynamic allocation, requires cleanup
  myStruct* dynamicStruct;
  dynamicStruct = malloc(sizeof(myStruct));
  if (dynamicStruct == NULL) {
    printf("Memory allocation error!\n");
    return (-1);
  } else {
    printf("Successfully allocated memory!\n");
  }

  dynamicStruct->i = 1;
  dynamicStruct->c = 'd';
  free(dynamicStruct);  // Release allocated memory
  dynamicStruct = NULL; // Somewhat common practise, though not 100% necessary
  return 0;
}

Now, if you need to create an array of dynamically allocated structs, and you've used them all up, and need more, you'd likely be best off with a slightly more complicated approach, like a dynamically allocated linked list of structs. A good example can be found in the "References" section below. Also, I've included a link to a somewhat related question I answered on memory allocation in C. It has some good examples that might also help clear up this topic for you.

References

1. C Linked List Data Structure Explained with an Example C Program, Accessed 2014-03-25, <http://www.thegeekstuff.com/2012/08/c-linked-list-example/>
2. Difference between declared string and allocated string, Accessed 2014-03-25, <https://stackoverflow.com/questions/16021454/difference-between-declared-string-and-allocated-string>

You are trying to allocate a array with the size of the pointer to the date struct instead of the actual size of the date struct.

Change date* to date:

array = malloc(size*sizeof(date));

Furthermore you don't need to allocate the day and year variables, because the malloc allocates them for you.

for (i=0; i<size; i++)
{
     array[i].month = calloc(MAX_MONTH_STR,sizeof(char));
     array[i].day = 0;
     array[i].year = 0;
}

Short answer: they are allocated with the order as they declared in the struct.

Example:

#include <stdio.h>
#include <string.h>

struct student 
{
    int id1;
    int id2;
    char a;
    char b;
    float percentage;
};

int main() 
{
    int i;
    struct student record1 = {1, 2, 'A', 'B', 90.5};

    printf("size of structure in bytes : %d\n", 
        sizeof(record1));

    printf("\nAddress of id1        = %u", &record1.id1 );
    printf("\nAddress of id2        = %u", &record1.id2 );
    printf("\nAddress of a          = %u", &record1.a );
    printf("\nAddress of b          = %u", &record1.b );
    printf("\nAddress of percentage = %u",&record1.percentage);

    return 0;
}

Output:

size of structure in bytes : 16 
Address of id1 = 675376768
Address of id2 = 675376772
Address of a = 675376776
Address of b = 675376777
Address of percentage = 675376780

The pictorial representation of above structure memory allocation is given below. This diagram will help you to understand the memory allocation concept in C very easily.

Further reading: check out here (also the source for the above example) for C – Structure Padding and Structure dynamic memory allocation in C.

The reason you are getting this problem is most likely the thing that @Steephen has already pointed out: (*sys)->c_array = malloc(num_challenges * sizeof(Activity));. It's most likely that the extra bracket was causing the error.