On Fri, Feb 8, 2019 at 7:36 AM Todd Chester via perl6-users < perl6-us...@perl.org> wrote:
> 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? > The "elegant" way I'd do it, is using unpack(): https://docs.perl6.org/routine/unpack It's experimental, so a declaration is needed, but Buf does Blob, so otherwise, it's straight to the point: $ perl6 -e 'use experimental :pack; my Buf $x=Buf.new(0xAE,0x5D,0x5C,0x72); say $x.unpack("L").base(0x10);' 725C5DAE $ Eirik
