Dynamically allocate memory for Array of Structs

Question

Here's what I'm trying to do:

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

struct myStruct {
    int myVar;
}

struct myStruct myBigList = null;

void defineMyList(struct myStruct *myArray)
{
     myStruct *myArray = malloc(10 * sizeof(myStruct));

     *myArray[0] = '42';
}

int main()
{
     defineMyList(&myBigList);
}

I'm writing a simple C program to accomplish this. I'm using the GNU99 Xcode 5.0.1 compiler. I've read many examples, and the compiler seems to disagree about where to use the struct tag. Using a struct reference inside the sizeof() command doesn't seem to recognize the struct at all.

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

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