How can i pass a structure to a function - c

Question

I have called a function with seven parameters from main
find_sync(es_data + (loop_count * size) + chunk_bytes_counter,
size - chunk_bytes_counter, &sync_index, &flag,
&sync_length, &chunk_bytes_counter, &total_bytes_counter);
in function.c:
void find_sync(char data[], size_t size, unsigned int *sync_index, int *flag, unsigned int *sync_length, unsigned int *chunk_bytes_counter, unsigned int *total_bytes_counter)
prototype in header file:
extern void find_sync(char data[], size_t size, unsigned int *sync_index, int *flag, unsigned int *sync_length, unsigned int *bytes_counter, unsigned int *total_bytes_counter);
Now, my question is, how can i declare all these 7 parameters in a structure, so that i can only pass one structure variable.

Answer 1

Begin by declaring the struct:
struct find_sync_parameters {
char* data;
size_t size;
unsigned int *sync_index;
int *flag;
unsigned int *sync_length;
unsigned int *bytes_counter;
unsigned int *total_bytes_counter;
}
Then change your function signature either to:
void find_sync(struct find_sync_parameters param)
Or to
void find_sync(struct find_sync_parameters *param)
In the first case the whole struct will be pushed onto the stack before transferring control to find_sync. On the second case only a pointer to the struct (stored elsewhere) will be pushed.
There are advantages and drawbacks in each one. When passing a pointer note that the function can change the contents (this can be positive: for returning values directly inside the struct; also can be negative: the caller cannot be sure if its data were changed or not). If the struct is too big (not your case), then pushing everything onto the stack can take a significant amount of time and become a performance hit.
Inside the function you use it either with '.' (dot, the first case) or '->' (arrow, the second case) operator.
To call it:
struct find_sync_parameters p = { ... };
find_sync(p); // first case
find_sync(&p); // second case
If you find it annoying to type struct find_sync_parameters everytime you can define a new type with typedef:
typedef struct find_sync_parameters find_sync_parameters;
Or in one line (struct and typedef definitions):
typedef struct find_sync_parameters {
...
} find_sync_parameters;
Or even without struct name (anonymous struct)
typedef struct {
...
} find_sync_parameters;
In this last case you cannot reference the struct itself inside the struct (the case, for example, with linked list nodes).

Answer 2

Just put in structure .
struct pars_t
{
char data[];
size_t size; unsigned int *sync_index;
int *flag; unsigned int *sync_length;
unsigned int *bytes_counter;
unsigned int *total_bytes_counter;
} pars;
and then call foo (pars_t par)

Answer 3

You can create a struct (and typedef it at the same time, to make it usable without saying "struct" every time) like so:
typedef struct _find_sync_str{
char* data;
size_t size;
unsigned int *sync_index;
int *flag;
unsigned int *sync_length;
unsigned int *bytes_counter;
unsigned int *total_bytes_counter;
} find_sync_str
Then you can list the function as:
void find_sync(find_sync_str f);

Answer 4

I am going to suggest:
struct find_sync_struct {
char* data;
size_t size;
unsigned int sync_index;
int flag;
unsigned int sync_length;
unsigned int bytes_counter;
unsigned int total_bytes_counter;
};
Change the input argument of find_sync to:
void find_sync(struct find_sync_struct* strPtr);
Call the function using:
struct find_sync_struct str;
// Set str.data to something suitable.
// ....
find_sync(&str);

Answer 5

Here is a simple example demonstrating how to pass a structure to a function:
#include<stdio.h>
#include<conio.h>
//-------------------------------------
struct Example
{
int num1;
int num2;
}s[3];
//-------------------------------------
void accept(struct Example *sptr)
{
printf("\nEnter num1 : ");
scanf("%d",&sptr->num1);
printf("\nEnter num2 : ");
scanf("%d",&sptr->num2);
}
//-------------------------------------
void print(struct Example *sptr)
{
printf("\nNum1 : %d",sptr->num1);
printf("\nNum2 : %d",sptr->num2);
}
//-------------------------------------
void main()
{
int i;
clrscr();
for(i=0;i<3;i++)
accept(&s[i]);
for(i=0;i<3;i++)
print(&s[i]);
getch();
}