Why doesn't @alignOf(*anyopaque) return 1?

Explain
1

Hello,

I was expecting @alignOf(*anyopaque) to return 1, but tried it with zig 0.14.0 on my machine and got 8.

const std = @import("std");

pub fn main() !void {
    std.debug.print("@alignOf(*anyopaque) == {d}", .{@alignOf(*anyopaque)});
}

I was expecting 1 because from my understanding:

  • If T is a type then, by default, @alignOf(T) == @alignOf(*T)
  • @alignOf(anyopaque) == 1
  • If an anyopaque pointer can point to any type, then it makes sense for it to have an alignment of 1
  • if you try to @ptrCast a *anyopaque you (can) get an error with a note saying " ‘*anyopaque’ has alignment ‘1’ "
2

This isn’t correct.

comptime {
    @compileLog(@alignOf(u8), @alignOf(*u8));
}

Compile Log Output:
@as(comptime_int, 1), @as(comptime_int, 8)
4 Likes
3

I’m going to assume you understand where alignment comes from and just answer:
a pointer has the same size as usize which varies by architecture, on x86_64 it’s 8 bytes. So it has an alignment of 8

2 Likes
4

Oh, I think I understand why I got confused:

For some reason, I thought @alignOf(*T) was the same as @typeInfo(*T).pointer.alignment.
But the latter returns the alignment of T, not *T, right?

Now it makes sense, pointers are aligned like usize, anyopaque represents an unknown type with unknown size, so its alignment is 1.

I wasn’t thinking straight, alright, thank you!