7 ms·
Yeah, after looking it up, it looks like it is basically only used as either field access or an 'infer operator', is that right? I thought it was used in four
by coxmi 3mo ago
Yeah, after looking it up, it looks like it is basically only used as either field access or an 'infer operator', is that right?
I thought it was used in four completely separate ways:
· normal struct field access
· anonymous struct definition
· field definition within structs (for reasons to do with the parser)
· an extra 'infer operator' for syntactic sugar
But there's no support for anonymous structs/fields, and all structs and fields require a type somewhere for it to be inferred. Which is why this is invalid zig:
const test = .{ .x = 0, .y = 1 };
(It would need the type to be specified in the called function definition, or inline when assigning)
Correct me if I'm wrong here! (And thank you)
- tremon 3mo agoI don't think that "infer operator" is a special case of field access, to me it feels like regular known-type elision similar to how C# and C++ use the var keyword if the data type can be inferred from the rhs expression: const Enum = enum {one, two, five}; const t: Enum = .one; // Enum.one, but the type was inferred from lhs std.debug.print("{t}\n", .{t}); Defining an anonymous struct is valid in zig; your example is only invalid because "test" is a reserved keyword. But you are correct that it reifies into a concrete type, and after initialization it doesn't coerce into other types because zig doesn't do structural typing: const anonymous = .{ .x = 0, .y = 1 }; std.debug.print("{}\n", .{@TypeOf(anonymous)}); // will output something like test_0__struct_45138 const Point = struct { x: i32, y: i32 }; const p1 = Point{ .x = 0, .y = 1 }; // valid, explicit struct literal const p2: Point = .{ .x = 0, .y = 1 }; // valid, anonymous struct will coerce to Point //const pt: Point = anonymous; // error: expected type 'test_0.Point', found 'test_0__struct_45138' And then there's fieldless anonymous structs aka tuples. I'm including them because they were used in the print statements above: const tuple = .{ 0, "1", true }; std.debug.print("{}\n", .{@TypeOf(tuple)}); // struct { comptime comptime_int = 0, comptime *const [1:0]u8 = "1", comptime bool = true }
- coxmi 3mo agoThanks for the detailed answer :) All this does for me is raise the question of why they chose to use the `.` for so many different uses. I'd be fine if it was just to infer the type, but it seems very overloaded.