Allocating Memory for a variable-sized array of Structs

[UncleRic]

Greetings:
I'm trying to get the ASCI C syntax correct here. I want to create memory space for a variable-size array of pointers to structs.

This is what I found in theScripts' archive:

> The following code:
>
> #include <stdio.h>
>
> // const int const_asize = 10;
> #define define_asize = 10;
>
> int array[define_asize] = {1,2,3,4,5,6,7,8,9,0};[/color]

You don't need the size. The compiler can count the initializers and
size the array for you :

int array[] = {1,2,3,4,5,6,7,8,9,0};

You can retreive the size by applying the definition of an array
(sequence of elements of the same size), hence :

size_t n = sizeof array / sizeof array[0];

or

size_t n = sizeof array / sizeof *array;

This is what I'm toying with:

#include <stdio.h>
#include <string.h>
 
typedef struct {
    char* name;  
    int   number;
} SomeSeq;
 
int main () {
    printf("Hello, World!\n");
    const unsigned n = 5;
    SomeSeq *myStructures[] = calloc(n,sizeof(SomeSeq));
    // ...
    free(myStructures);
    // ...
    return 0;
}
 

I'm getting the following compiler errors:

main.c:12: warning: incompatible implicit declaration of built-in function 'calloc'
main.c:12: error: invalid initializer
 

What's wrong with this syntax?
... I even tried to do a (SomeSeq *) cast of calloc, but doesn't correct the problems.

Regards,

Ric.

[RRick]

Your problem is on line 12, where you have defined too many levels of indirections for your pointer (i.e. loose the [ ] and it should work).

With C arrays, each * and [ ] defines another dimension (or level of indirection). You have defined an array of pointers where each thing they point to is an array of SomeSeq. You really only want a single array of SomeSeq defined. That is what calloc is expecting.

[UncleRic]

Your problem is on line 12, where you have defined too many levels of indirections for your pointer (i.e. loose the [ ] and it should work).

With C arrays, each * and [ ] defines another dimension (or level of indirection). You have defined an array of pointers where each thing they point to is an array of SomeSeq. You really only want a single array of SomeSeq defined. That is what calloc is expecting.

Thanks for the clarity. So the following should be okay:

const unsigned n = 5;
SomeSeq *myStructures = calloc(n,sizeof(SomeSeq));
 

or...

const unsigned n = 5;
SomeSeq *myStructures = malloc(sizeof(SomeSeq)*n);
 

This would give me an array of pointers to 5 structures on the heap.
...I think.

Now...

To access each of the elements, should I do something like:

const unsigned n = 5;
SomeSeq *myStructures = malloc(sizeof(SomeSeq)*n);
myStuctures[0]->name='Ric';
...
...
free(myStructures);
...
 

I believe line #3 above is incorrect. What would be the correct syntax or accessing a Struct member via an array of struct pointers?

... my shared Wi-Fi signal is waning... hence in a rush.

Thanks for your help!!

Regards,
Ric.

[weaknessforcats]

You may be making this too hard.

For an array of pointers, malloc() or calloc() will return the address of the array. That means you need the address of a pointer to receive the return from malloc() or calloc():

SomeSeq** ptr = (SomeSeq**) malloc(5 * sizeof(SomeSeq*));