If you're depending on sizeof(yourstruct)
being 42 bytes, you're about to be bitten by a world of non-portable assumptions. You haven't said what this is for, but it seems likely that the endianness of the struct contents matters as well, so you may also have a mismatch with the x86 there too.
In this situation I think the only sure-fire way to cope is to use unsigned char[42]
in the parts where it matters. Start by writing a precise specification of exactly what fields are where in this 42-byte block, and what endian, then use that definition to write some code to translate between that and a struct you can interact with. The code will likely be either all-at-once serialisation code (aka marshalling), or a bunch of getters and setters.
The examples given in the book are highly dependent on the used compiler and computer architecture. If you test them in your own program you may get totally different results than the author. I will assume a 64-bit architecture, because the author does also, from what I've read in the description.
Lets look at the examples one by one:
ReallySlowStruct
IF the used compiler supports non-byte aligned struct members, the start of "d" will be at the seventh bit of the first byte of the struct. Sounds very good for memory saving. The problem with this is, that C does not allow bit-adressing. So to save newValue to the "d" member, the compiler must do a whole lot of bit shifting operations: Save the first two bits of "newValue" in byte0, shifted 6 bits to the right. Then shift "newValue" two bits to the left and save it starting at byte 1. Byte 1 is a non-aligned memory location, that means the bulk memory transfer instructions won't work, the compiler must save every byte at a time.
SlowStruct
It gets better. The compiler can get rid of all the bit-fiddling. But writing "d" will still require writing every byte at a time, because it is not aligned to the native "int" size. The native size on a 64-bit system is 8. so every memory address not divisable by 8 can only be accessed one byte at a time. And worse, if I switch off packing, I will waste a lot of memory space: every member which is followed by an int will be padded with enough bytes to let the integer start at a memory location divisable by 8. In this case: char a and c will both take up 8 bytes.
FastStruct
this is aligned to the size of int on the target machine. "d" takes up 8 bytes as it should. Because the chars are all bundled at one place, the compiler does not pad them and does not waste space. chars are only 1 byte each, so we do not need to pad them. The complete structure adds up to an overall size of 16 bytes. Divisable by 8, so no padding needed.
In most scenarios, you never have to be concerned with alignment because the default alignment is already optimal. In some cases however, you can achieve significant performance improvements, or memory savings, by specifying a custom alignment for your data stuctures.
In terms of memory space, the compiler pads the structure in a way that naturally aligns each element of the structure.
struct x_
{
char a; // 1 byte
int b; // 4 bytes
short c; // 2 bytes
char d; // 1 byte
} bar[3];
struct x_
is padded by the compiler and thus becomes:
// Shows the actual memory layout
struct x_
{
char a; // 1 byte
char _pad0[3]; // padding to put 'b' on 4-byte boundary
int b; // 4 bytes
short c; // 2 bytes
char d; // 1 byte
char _pad1[1]; // padding to make sizeof(x_) multiple of 4
} bar[3];
Source: https://learn.microsoft.com/en-us/cpp/cpp/alignment-cpp-declarations?view=vs-2019
Best Answer
At least on most machines, a type is only ever aligned to a boundary as large as the type itself [Edit: you can't really demand any "more" alignment than that, because you have to be able to create arrays, and you can't insert padding into an array]. On your implementation,
short
is apparently 2 bytes, andint
4 bytes.That means your first struct is aligned to a 2-byte boundary. Since all the members are 2 bytes apiece, no padding is inserted between them.
The second contains a 4-byte item, which gets aligned to a 4-byte boundary. Since it's preceded by 6 bytes, 2 bytes of padding is inserted between
v3
andi
, giving 6 bytes of data in theshort
s, two bytes of padding, and 4 more bytes of data in theint
for a total of 12.