Question

我想转换十六进制字符串说

[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}

进入字节数组

4A060EF1953BD9905B63CAA937C88DDA6482999FE11A3BB6FC68C0D27B0121DD

和反向，有人可以建议吗？

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

取决于字节数组的含义：

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

反之亦然：

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

或者：

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

反之亦然：

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

如何使用Perl将十六进制字符串转换为字节数组

2 个答案: