If you have a new enough version of Rakudo:
my Buf $x=Buf.new(0xAE,0x5D,0x5C,0x72);
my int32 $i = $x.read-int32(0,LittleEndian);
say $i.base(16);
# 725C5DAE
On Fri, Feb 8, 2019 at 12:35 AM Todd Chester via perl6-users
<perl6-us...@perl.org> wrote:
>
> Hi All,
>
> I am dealing with a Buf what includes 32 bit integers, but
> they are entered somewhat backwards as view with hexedit:
>
> AE 5D 5C 72 represents the number 725C5DAE
>
> This is what I have come up with to convert this type of
> number in a buffer to and integer
>
> $ p6 'my Buf $x=Buf.new(0xAE,0x5D,0x5C,0x72); my int32 $i=$x[3] +< 0x18
> + $x[2] +< 0x10 + $x[1] +< 0x08 + $x[0]; say $x; say $i.base(0x10);'
>
> Buf:0x<ae 5d 5c 72>
> 725C5DAE
>
>
> Is there a more "elegant" way to do this?
>
> Many thanks,
> -T
