Dynamically allocate memory for Array of Structs

Question

Here\'s what I\'m trying to do:

#include 
#include 

struct myStruct {
    int myVar;
}

struct myStruct myBigList = null;

vo         


        

Answer 1

There are a few errors in your code. Make it:

struct myStruct *myBigList = NULL; /* Pointer, and upper-case NULL in C. */

/* Must accept pointer to pointer to change caller's variable. */
void defineMyList(struct myStruct **myArray)
{
     /* Avoid repeating the type name in sizeof. */
     *myArray = malloc(10 * sizeof **myArray);

     /* Access was wrong, must use member name inside structure. */
     (*myArray)[0].myVar = 42;
}

int main()
{
     defineMyList(&myBigList);
     return 0; /* added missing return */
}

Basically you must use the struct keyword unless you typedef it away, and the global variable myBigList had the wrong type.

Answer 2

    sizeof(struct myStruct)

or

    typedef struct myStruct myStrut;
    sizeof(myStruct)

Answer 3

Shouldn't the following statement

myArray[0].myVar = '42'; 

be this?

(*myArray)[0].myVar = 42;

myvar is an integer.

Answer 4

In order to work for all 10 elements for that array the line:

myArray[0].myVar = '42';

should be:

(*myArray)[0].myVar = '42';

Answer 5

This is because struct name is not automatically converted into a type name. In C (not C++) you have to explicitly typedef a type name.

Either use

struct myStruct instance;

when using the type name OR typedef it like this

typedef struct {
    int myVar;
} myStruct;

now myStruct can simply be used as a type name similar to int or any other type.

Note that this is only needed in C. C++ automatically typedefs each struct / class name.

A good convention when extending this to structs containing pointers to the same type is here