ANSI C / C89 struct to int array[]

Question

Is there an efficient way to assign/convert a struct to array and vice-versa?

The struct I have is as follows:

struct A {
    int x, y;
}

struct B {
    struct A start;
    struct A end;
}

Basically they contain xy coordinates for start and end positions.

I need to assign them efficiently however currently I can only do this

/* sample code */
struct B b;
b.start.x = arr[i];
b.start.y = arr[i];
b.end.x = arr[i];
b.end.y = arr[i];


/* I can't do this in ANSI C / C89 as compound literals only allow constants */
b = (struct B) {(struct A) {arr[0], arr[1]}, (struct A) {arr[2], arr[3]}};

I can use compound literals as well but it gives me a warning in gcc when I compile with flags -Wall -pedantic -ansi

Is there a way to reduce those 4 lines of assignment to just one without getting a warning with the flags mentioned above.

Regards

Edit: fixed compound literal syntax

Answer 1

struct A
{
    int x, y;
};

struct B
{
    struct A start;
    struct A end;
};

void InitA(struct A* s, int x, int y)
{
    s->x = x;
    s->y = y;
}

void InitB(struct B* s, int x1, int y1, int x2, int y2)
{
    InitA(&s->start, x1, y1);
    InitA(&s->end, x2, y2);
}

void InitBFromArray(struct B* s, int *a)
{
    InitB(s, a[0], a[1], a[2], a[3]);
}

int main()
{
    int a[] = { 1, 2, 3, 4 };
    struct B s;
    InitBFromArray(&s, a);
}