Nik*_*l S 2 python perl struct unpack
我有一个二进制文件,其中包含4 KB的头信息,然后是28个字节的数据,然后是24个字节,我想要读取.我如何循环每24和28字节并读取(或提取)那28和24字节的每个前8字节数据..在python我做了类似的事情.不知道如何做可变长度
import sys
import struct
f = open(sys.argv[1],"rb")
f.seek(4096)
byte = f.read(28)
while byte != "":
ticks = struct.unpack("<ll",byte[:8]) #not sure how to read 8 bytes
byte = f.read(28)
f.close()
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这是标题后的样子.
Length
(bytes) Field Name
8 TS_INCR
4 SEQID
2 OP
2 LUN
4 NBLKS
8 LBA
Length
(bytes) Field Name
8 TS_INCR
4 SEQID
2 OP
2 LUN
4 LATENCY_TICKS
2 HOST_ID
2 HOST_LUN
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如果你们可以帮忙解决这个问题.Python或PERL并不重要.谢谢!!!!
您正在阅读的数据的字节顺序非常重要.你似乎正在解压缩8个八位组,因为两个长点以小端顺序存储.您确定它不是一个64位数量(这会使格式q或Q格式更合适)吗?不幸的是,我在32位机器上,所以我perl不支持Q.
但是,以下内容应指向正确的方向:
#!/usr/bin/env perl
use strict; use warnings;
use autodie;
use Fcntl qw(:seek);
use List::Util qw( sum );
my ($input_file) = @ARGV;
die "Need input file\n" unless defined $input_file;
my $HEADER_SIZE = 4_096;
my @typedef = (
{
fields => [
qw(
TS_INCR_LO
TS_INCR_HI
SEQID
OP
LUN
NBLKS
LBA_LO
LBA_HI
)
],
tmpl => 'LLLSSLLL',
start => 0,
size => 28,
},
{
fields => [
qw(
TS_INCR_LO
TS_INCR_HI
SEQID
OP
LUN
LATENCY_TICKS
HOST_ID
HOST_LUN
)
],
tmpl => 'LLLSSLSS',
start => 28,
size => 24,
},
);
open my $input, '<:raw', $input_file;
seek $input, $HEADER_SIZE, SEEK_SET;
my $BLOCK_SIZE = sum map $_->{size}, @typedef;
read $input, my($buffer), $BLOCK_SIZE;
my @structs;
for my $t ( @typedef ) {
my %struct;
@struct{ @{ $t->{fields}} } = unpack(
$t->{tmpl},
substr($buffer, $t->{start}, $t->{size})
);
push @structs, \%struct;
}
use Data::Dumper;
print Dumper \@structs;
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