NativeCall中的位字段

Question

我正在尝试为C图创建Perl6绑定，其中一个结构为它的一些属性设置了位字段，其值小于8。我应该如何在我的模块中表示它呢？

我尝试过使用is nativesize(x)特性定义自定义类型，但是CStructs只支持8位宽的倍数类型。

C示例代码：

struct Agtag_s {
    unsigned objtype:2;
}

我试过的是：

my native objtype is repr('P6int') is Int is nativesize(2) is export { }
class Agtag is repr('CStruct') is export {
    has objtype $.object-type;
}

试图在该代码中使用我的模块失败了，出现了以下错误消息：CStruct only supports native types that are a multiple of 8 bits wide (was passed: 2)

Answer 1

下面是一个例子。我假设函数use_struct()是在库libslib中定义的：

#include <stdio.h>

struct Agtag_s {
    unsigned objtype:2;
    unsigned footype:4;
    unsigned bartype:6;
};

void use_struct (struct Agtag_s *s) {
    printf("sizeof(struct Agtag_s): %ld\n", sizeof( struct Agtag_s ));
    printf("objtype = %d\n", s->objtype);
    printf("footype = %d\n", s->footype);
    printf("bartype = %d\n", s->bartype);
    s->objtype = 3;
    s->footype = 13;
    s->bartype = 55;
}

然后在Perl 6中：

use v6;
use NativeCall;

class Agtag is repr('CStruct') is export {
    has int32 $.bitfield is rw;
}

sub use_struct(Agtag $s is rw) is native("./libslib.so") { * };

my $s = Agtag.new();
my $objtype = 1;
my $footype = 7;
my $bartype = 31;
$s.bitfield = $objtype +| ($footype +< 2 ) +| ($bartype +< 6);
say "Calling library function..";
say "--------------------------";
use_struct( $s );
say "After call..";
say "------------";
say "objtype = ", $s.bitfield +& 3;
say "footype = ", ($s.bitfield +> 2) +& 15;
say "bartype = ", ($s.bitfield +> 6) +& 63;

输出

Calling library function..
--------------------------
sizeof(struct Agtag_s): 4
objtype = 1
footype = 7
bartype = 31
After call..
------------
objtype = 3
footype = 13
bartype = 55