declare pointer and allocate memory for an array typedef

[Amar Akshat]

Hi, In my attempt to use a library called PJSIP, I came across a definiton for a typedef as following,

typedef long some_type[64];

Now, I want to declare a pointer and allocate memory for this in a thread.
I DONT want to declare and automatically allocate memory like this,

some_type a;

Rather I want to do something like this,

some_type* a = malloc(sizeof(*a));

and pass 'a' to a function, with a definition like: void function(some_type a);

however, the effective sizeof allocated memory shall be 256, but its only 4, which is nothing but the sizeof of a pointer.

How can I possibly declare and allocate a memory for a variable of type "some_type".

Thanks

Amar

[laserlight]

however, the effective sizeof allocated memory shall be 256, but its only 4, which is nothing but the sizeof of a pointer.

Are you sure? How did you check?

[Salem]

Declaring a typedef of an array doesn't change the underlying implementation that all arrays are passed as pointers. You might not have any [] or * in the function declaration (as written), but they're still there.

Eg.

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

typedef long some_type[64];

void function(some_type a) {
  printf("%ld\n", sizeof(a) );
}

int main(void) {
  some_type a;
  some_type* pa = malloc(sizeof(*pa));
  long arr[64];
  function(a);
  function(*pa);
  function(arr);
  return 0;
}

A typedef is just an alias for some other type. If you want, you can write out the long form and get exactly the same results.

If you want to pass arrays "by value" (only really worth doing for small arrays), then you can hide the array in a struct, like

struct sometype {
    long arr[64];
};

Remember to keep an eye on the size of your structs. Passing structs by value involves memcpy, and that will get expensive if you're doing it a lot with large structures.