在Perl中将二进制数据拆分为字节数组

3

我基本上想将一个二进制字符串转换为字节数组/列表(以便索引,并避免使用 substr,因为它的语法使事情变得复杂),然后我想出了以下的MWE

#!/usr/bin/env perl

use warnings;
use strict;

# Use open ':raw';      # Unknown PerlIO layer class ':raw'
use open IO => ':raw';

binmode(STDIN);
binmode(STDOUT);

# Create original 8-bit byte array/list
my @atmp = (0x80, 0x23, 0x14, 0x0d, 0x0a, 0x00, 0x00, 0x80, 0x43, 0x00, 0x00);

# Make a copy of portion
my @atmp2 = (0) x 2;
@atmp2[0..1] = @atmp[7..8];

# Print output
print "Copied atmp2 contents as hex: " . join(", ", unpack("H2"x2, pack("C"x2,@atmp2))) . "\n";
print "Copied atmp2 as ushort (16bit) int: " . unpack("S", pack("C"x2, @atmp2));
# doublecheck value by routing through printf with format specifier:
printf(" [%d]\n", unpack("S", pack("C"x2, @atmp2)));


# Now, the same data as string:
my $indata = "\x80\x23\x14\x0d\x0a\x00\x00\x80\x43\x00\x00";

# Create byte array (by converting string $indata to array/list with `split`)
my @btmp = split('',$indata);
print "lastindex: " . $#btmp . "\n";

# Make a copy of portion
my @btmp2 = (0) x 2;
@btmp2[0..1] = @btmp[7..8];

# Print output
print "Copied btmp2 contents as hex: " . join(", ", unpack("H2"x2, pack("C"x2,@btmp2))) . "\n";
print "Copied btmp2 as ushort (16bit) int: " . unpack("S", pack("C"x2, @btmp2));
# doublecheck value by routing through printf with format specifier:
printf(" [%d]\n", unpack("S", pack("C"x2, @btmp2)));

运行这段代码将会得到以下结果:
$ perl test.pl
Copied atmp2 contents as hex: 80, 43
Copied atmp2 as ushort (16bit) int: 17280 [17280]
lastindex: 10
Argument "M-\0" isn't numeric in pack at test.pl line 38.
Argument "C" isn't numeric in pack at test.pl line 38.
Copied btmp2 contents as hex: 00, 00
Copied btmp2 as ushort (16bit) int: 0 [0]

我该如何让第二部分(btmp2)与第一部分(atmp2)的行为相同?

1个回答

7
原来,在使用split时,它确实会创建一个包含与原始字符串相同字节的数组;但是,它似乎还将生成的数组标记为“文本”,因此进一步处理会导致“参数不是数字”的错误。
答案很简单,只需将split行替换为使用unpack的行即可:
- my @btmp = split('',$indata);
+ my @btmp = unpack('C*',$indata);

...之后,一切都按预期工作(两个打印输出均相同)。有趣的是,在这两种情况下,“lastindex”(对于从字符串派生的数组)将显示为10(这使我认为 binmode 可能有问题,这就是为什么代码中有所有这些语句的原因)。


1
好的,它确实帮了我,谢谢!我简直不敢相信回答这个简单问题有多么困难,但我找不到任何一个清晰解释的Perl文档、手册或教程... - Tobias J

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