将具有struct类型的值传递给C中的函数

Question

typedef struct {
        nat id;
        char *data;
        } element_struct;

typedef element_struct * element;

void push(element e, queue s) {
        nat lt = s->length;
        if (lt == max_length - 1) {
                printf("Error in push: Queue is full.\n");
                return;
        }
        else {
                s->contents[lt] = e;
                s->length = lt + 1;
        }
}
int main () {
         push(something_of_type_element, s);
}

How would i go about formatting " something_of_type_element "?

Thanks

Notes: nat is the same as int

Answer 1

How about:

element elem = malloc(sizeof(element_struct));
if (elem == NULL) {
    /* Handle error. */
}

elem->id = something;
elem->data = something_else;

push(elem, s);

Note that there's lots of memory management missing here...

Answer 2

Like this:

element_struct foo = { 1, "bar" };
push(&foo, s);

If you have a C99 compiler you can do this:

element_struct foo = {
    .id = 1,
    .data = "bar"
};
push(&foo, s);

Note that the data in the structure must be copied if it needs to live longer than the scope in which it was defined. Otherwise, memory can be allocated on the heap with malloc (see below), or a global or static variable could be used.

element_struct foo = malloc(sizeof (element_struct));

foo.id = 1;
foo.data = "bar";
push(foo, s);