使用nativeCall将cstruct中的内联carray传递给共享库

这是一个后续问题 How to declare native array of fixed size in Perl 6? “。

在这个问题中,讨论了如何将固定大小的数组合并到 CStruct . 在 this answer 建议使用 HAS 嵌入 CArray 在 C结构 . 当我测试这个想法时,我遇到了一些奇怪的行为,这些行为在问题下面的评论部分无法解决,所以我决定把它写成一个新问题。这是我的C测试库代码:

滑动C :

#include <stdio.h>

struct myStruct
{
    int A; 
    int B[3];
    int C;
};

void use_struct (struct myStruct *s) {
    printf("sizeof(struct myStruct): %ld\n", sizeof( struct myStruct ));
    printf("sizeof(struct myStruct *): %ld\n", sizeof( struct myStruct *));
    printf("A = %d\n", s->A);
    printf("B[0] = %d\n", s->B[0]);
    printf("B[1] = %d\n", s->B[1]);
    printf("B[2] = %d\n", s->B[2]);
    printf("C = %d\n", s->C);
}

要从中生成共享库,我使用了:

gcc -c -fpic slib.c
gcc -shared -o libslib.so slib.o

然后,Perl6代码:

P.P6 :

use v6;
use NativeCall;

class myStruct is repr('CStruct') {
    has int32 $.A is rw;
    HAS int32 @.B[3] is CArray is rw;
    has int32 $.C is rw;
}

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

my $s = myStruct.new();
$s.A = 1;
$s.B[0] = 2;
$s.B[1] = 3;
$s.B[2] = 4;
$s.C = 5;
say "Expected size of Perl 6 struct: ", (nativesizeof(int32) * 5);
say "Actual size of Perl 6 struct: ", nativesizeof( $s );
say 'Number of elements of $s.B: ', $s.B.elems;
say "B[0] = ", $s.B[0];
say "B[1] = ", $s.B[1];
say "B[2] = ", $s.B[2];
say "Calling library function..";
say "--------------------------";
use_struct( $s );

脚本的输出是:

Expected size of Perl 6 struct: 20
Actual size of Perl 6 struct: 24
Number of elements of $s.B: 3
B[0] = 2
B[1] = 3
B[2] = 4
Calling library function..
--------------------------
sizeof(struct myStruct): 20
sizeof(struct myStruct *): 8
A = 1
B[0] = 0         # <-- Expected 2
B[1] = 653252032 # <-- Expected 3
B[2] = 22030     # <-- Expected 4
C = 5

问题 :

• 为什么? nativesizeof( $s ) 给出24(而不是预期值20)?

• 为什么数组的内容是 B 在从C函数打印时结构是否与预期不符?

注释 :

我使用的是Ubuntu 18.04和Perl6 Rakudo版本2018.04.01,但也使用了版本2018.05进行了测试。