Given this struct:
struct PipeShm
{
int init;
int flag;
sem_t *mutex;
char * ptr1;
char * ptr2;
int status1;
int status2;
in
Okay I got it :
static struct PipeShm myPipeSt = {.init = 0 , .flag = FALSE , .mutex = NULL , .ptr1 = NULL , .ptr2 = NULL ,
.status1 = -10 , .status2 = -10 , .semaphoreFlag = FALSE };
static struct PipeShm * myPipe = &myPipeSt;
First you need to allocate memory for the pointer as below:
myPipe = malloc(sizeof(struct PipeShm));
Then, you should assign values one by one as below:
myPipe->init = 0;
myPipe->flag = FALSE;
....
Please note that for each individual pointer inside the structure, you need allocate memory seperately.
First initialize the struct (static struct PipeShm myPipe = {...
). Then take the address
struct PipeShm * pMyPipe = &myPipe;
Is it possible to declare a pointer to a struct and use initialization with it ?
Yes.
const static struct PipeShm PIPE_DEFAULT = {.init = 0 , .flag = FALSE , .mutex = NULL , .ptr1 = NULL , .ptr2 = NULL ,
.status1 = -10 , .status2 = -10 , .semaphoreFlag = FALSE };
static struct PipeShm * const myPipe = malloc(sizeof(struct PipeShm));
*myPipe = PIPE_DEFAULT;
you have to build that struct by hand, and then make a pointer pointing to that.
either
static struct PipeShm myPipe ={};
static struct PipeShm *pmyPipe = &myPipe;
or
static struct PipeShm *myPipe = malloc();
myPipe->field = value;
You can do it like so:
static struct PipeShm * myPipe = &(struct PipeShm) {
.init = 0,
/* ... */
};
This feature is called a "compound literal" and it should work for you since you're already using C99 designated initializers.
Regarding the storage of compound literals:
6.5.2.5-5
If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.