On Monday, 30 July 2012 at 14:19:56 UTC, Andrej Mitrovic wrote:
It says for std.bitmanip.bitfields:
"The sum of all bit lengths in one bitfield instantiation must
be
exactly 8, 16, 32, or 64."
It has a static assert to verify this. But why does this
limitation exist?
I was thinking of cases where I'm wrapping a C++ POD struct
which has
bitfields and is packed on a 1-byte boundary. GCC example:
#include <stdio.h>
#pragma pack(1)
struct Foo
{
unsigned int bits1 : 1;
unsigned int bits2 : 2;
};
int main()
{
printf("%d\n", sizeof(Foo));
return 0;
}
I can't translate this to D using the bitfields mixin since it
requires a multiple of a byte, 3 bits isn't a byte:
import std.stdio;
import std.bitmanip;
align(1) struct Foo
{
mixin(bitfields!(
uint, "bits1", 1,
uint, "bits2", 2
));
}
void main() { writeln(Foo.sizeof); }
std\bitmanip.d(151): Error: static assert "Field widths must
sum to
8, 16, 32, or 64"
Is this a language limitation? I guess because 'byte' is the
smallest
legal type in D this is what the bitfields mixin uses
internally?
I think you are supposed to explicitly pad the field yourself:
----
align(1) struct Foo
{
mixin(bitfields!(
uint, "bits1", 1,
uint, "bits2", 2,
uint, "", 5
));
}
----
The rationale (I *think*) Is that bitfields wants to know how big
it should be: While YOU know you want to put it into a "Foo"
object that is 1 byte big, bitfields has no idea.
Either that, or having the developer provide this by hand made
the implementation easier enough to justify the burden?
Even then, I think it would have been more convenient to specify
the total size as a first argument, rather than manually padding
at the end.
Either way, nothing world shattering for me :D
