perl - Perl を使用して 16 進文字列をバイト配列に変換する方法

Question

16 進数の文字列を変換したい

[0] =0x4A、[1] =0x06、[2] =0x0E、[3] =0xF1、[4] =0x95、[5] =0x3B、[6] =0xD9、[7] =0x90、[8 ] =0x5B ,[9] =0x63 ,[10]=0xCA ,[11]=0xA9 ,[12]=0x37 ,[13]=0xC8 ,[14]=0x8D ,[15]=0xDA ,[16]= 0x64、[17]=0x82、[18]=0x99、[19]=0x9F、[20]=0xE1、[21]=0x1A、[22]=0x3B、[23]=0xB6、[24]=0xFC、 [25]=0x68 ,[26]=0xC0 ,[27]=0xD2 ,[28]=0x7B ,[29]=0x01 ,[30]=0x21 ,[31]=0xDD }

としてバイト配列に

4A060EF1953BD9905B63CAA937C88DDA6482999FE11A3BB6FC68C0D27B0121DD

逆に、誰かが提案してもらえますか?

Answer 1

$str = '[0] =0x4A ,[1] =0x06 ,[2] =0x0E ,[3] =0xF1 ,[4] =0x95 ,[5] =0x3B ,[6] =0xD9 ,[7] =0x90 ,[8] =0x5B ,[9] =0x63 ,[10]=0xCA ,[11]=0xA9 ,[12]=0x37 ,[13]=0xC8 ,[14]=0x8D ,[15]=0xDA ,[16]=0x64 ,[17]=0x82 ,[18]=0x99 ,[19]=0x9F ,[20]=0xE1 ,[21]=0x1A ,[22]=0x3B ,[23]=0xB6 ,[24]=0xFC ,[25]=0x68 ,[26]=0xC0 ,[27]=0xD2 ,[28]=0x7B ,[29]=0x01 ,[30]=0x21 ,[31]=0xDD';
$str =~ s/\s*,?\[.*?\]\s*=0x//gi;
print $str, "\n";
$str =~ s/([0-9A-F]{2})/0x$1, /gi;
print $str, "\n";

Answer 2

バイト配列の意味に応じて：

my $dump = '[0] =0x4A ...';
my $bytes = pack 'H*', join '', $dump =~ /0x(..)/sg;

逆に：

my $bytes = "\x4A\06...";
my @bytes = unpack 'C*', $bytes;
my $dump = join ', ', map sprintf("[%d] = 0x%02X", $_, $bytes[$_]), 0..$#bytes;

または：

my $dump = '[0] =0x4A ...';
my @bytes = map hex, $dump =~ /0x(..)/sg;

逆に：

my @bytes = (0x4A, 0x06, ...);
my $dump = join ', ', map sprintf("[%d] = 0x%02X", $_, $bytes[$_]), 0..$#bytes;