How to get used bytes of a packed struct?

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1

Consider that I have a packed struct with underlying backing integer u112 (14 bytes).

On my system, this packed struct will align to u128 (16 bytes).

How do I obtain a slice of only the bytes used by the packed struct without the padding bytes?

2

@bitSizeOf(T) returns the actual bits used by a packed struct.
To get the bytes you need to ceiling divide by 8.

3

Which bytes are the paddings bytes? Is it always highest address or most significant bytes?

4

The first member of the packed struct is in the first bytes or least significant bytes.
The padding bytes are the remaining last bytes or most significant bytes.

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5

This should help answer your question:

const P112 = packed struct(u112) {
    a: u64 = 0, // eight
    b: u32 = 0xaaaaaaaa, // four (twelve)
    c: u8,
    d: u8,
};

test "packed unpacking" {
    std.debug.print("size of P112: {d}\n", .{@sizeOf(P112)});
    std.debug.print("bit size of P112: {d}\n", .{@bitSizeOf(P112)});
    const a_p112 = P112{ .c = 0x11, .d = 0xcc };
    const a_bytes = std.mem.asBytes(&a_p112);
    std.debug.print("byte length of cast P112: {d}\n", .{a_bytes.len});
    const a_real_bytes = a_bytes[0 .. 112 / 8];
    try std.testing.expectEqual(0xcc, a_real_bytes[a_real_bytes.len - 1]);
}
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6

Im not sure these solutions will work on big endian systems.

On a big endian system, the least significant bytes have the highest address (and thus are the last elements in a slice)

7

packed struct

Unlike normal structs, packed structs have guaranteed in-memory layout:

  • Fields remain in the order declared, least to most significant.
  • There is no padding between fields.
  • …

In big endian systems the order of the fields remain the same (least to most significant), but the order of bytes in fields change.

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