On 6/24/17 10:32 AM, Honey wrote:
On Saturday, 24 June 2017 at 12:41:47 UTC, Steven Schveighoffer wrote:
Any padding bits between fields should be 0 as long as the struct is
initialized (i.e. as long as you don't do Struct s = void).
Padding bits after the fields I assume would be 0, but I don't know if
this is defined. It's possible the compiler doesn't consider those
bits to be part of the struct, and just there for alignment.
There is no spec for this, but I know that when the compiler has to
fill gaps with something it chooses 0.
Thanks. Your answer has generated more questions. ;-)
Let's say, I have a struct S of size n with m bits of padding at the
end. How can I find m?
Is it possible to provide a facility Pad such that for any struct T,
Pad!T is a struct that mimics T but contains explicit instead of
implicit padding? E.g.
struct Foo
{
ubyte b;
double d;
int i;
}
struct Pad!Foo
{
ubyte b;
ubyte[7] __padding_0;
double d;
int i;
ubyte[4] __padding_1;
}
Of course it's possible. Best thing I can think of is using offsetof and
sizeof:
string generatePad(T)()
{
string result = "";
size_t lastIndex = 0;
foreach(x; T.tupleof)
{
// not sure about offsetof working like this, you may need to do
// tupleof with index
auto padSpace = x.offsetof - lastIndex;
if(padSpace > 0) result ~= "ubyte[" ~ padSpace ~ "] __padding__"
~ padIndex ~ ";";
result ~= typeof(x).stringof ~ " " ~ x.stringof ~ ";";
lastIndex = x.offsetof + x.sizeof;
}
return result;
}
This needs a bit of work, but you get the idea. I'm not sure all the
properties work correctly there. Then you mixin the result of that
function to get your members. Adding the ending padding would involve
comparing lastIndex to the sizeof T.
-Steve
