// Dot-shorthand `.{ child = d }` for a struct whose first field is a protocol
// value, used as the argument to `List(Container).append` from two distinct
// container types. Exercises the cross-callsite path of dot-shorthand inference.

#import "modules/std.sx";

Drawable :: protocol {
    draw :: () -> i32;
    name :: () -> string;
    layout :: (x: i32) -> i32;
    handle :: (event: i32) -> bool;
}

Circle :: struct { radius: i32; }
impl Drawable for Circle {
    draw :: (self: *Circle) -> i32 { self.radius }
    name :: (self: *Circle) -> string { "circle" }
    layout :: (self: *Circle, x: i32) -> i32 { x + self.radius }
    handle :: (self: *Circle, event: i32) -> bool { event > 0 }
}

Square :: struct { side: i32; }
impl Drawable for Square {
    draw :: (self: *Square) -> i32 { self.side * self.side }
    name :: (self: *Square) -> string { "square" }
    layout :: (self: *Square, x: i32) -> i32 { x + self.side }
    handle :: (self: *Square, event: i32) -> bool { event > 1 }
}

Rect :: struct {
    x: f32;
    y: f32;
    w: f32;
    h: f32;
    zero :: () -> Rect { Rect.{ x = 0.0, y = 0.0, w = 0.0, h = 0.0 } }
}

Container :: struct {
    child: Drawable;
    computed_frame: Rect = .zero();
}

// Two different structs, each with List(Container), both calling .append(.{...})
// This mirrors VStack/HStack in the game.

StackA :: struct {
    children: List(Container);

    add :: (self: *StackA, d: Drawable) {
        // BUG: `.{ child = d }` causes LLVM error when 2+ structs do this
        self.children.append(.{ child = d });
    }
}

StackB :: struct {
    children: List(Container);

    add :: (self: *StackB, d: Drawable) {
        // BUG: second struct doing `.{ child = d }` triggers the error
        self.children.append(.{ child = d });
        // FIX: explicit `Container.{ child = d }` works
        // self.children.append(Container.{ child = d });
    }
}

main :: () -> void {
    c := Circle.{ radius = 42 };
    s := Square.{ side = 5 };

    a : StackA = .{};
    a.add(c);
    print("StackA: draw={}\n", a.children.items[0].child.draw());

    b : StackB = .{};
    b.add(s);
    print("StackB: draw={}\n", b.children.items[0].child.draw());

    print("OK\n");
}