sha碰撞,MD5碰撞实现，#!/usr/local/bin/perl
# It was noted that Intel IA-32 C compiler generates code which
# performs ~30% *faster* on P4 CPU than original *hand-coded*
# SHA1 assembler implementation. To address this problem (and
# prove that humans are still better than machines:-), the
# original code was overhauled, which resulted in following
# performance changes:
#
# compared with original compared with Intel cc
# assembler impl. generated code
# Pentium -25% +37%
# PIII/AMD +8% +16%
# P4 +85%(!) +45%
#
# As you can see Pentium came out as looser:-( Yet I reckoned that
# improvement on P4 outweights the loss and incorporate this
# re-tuned code to 0.9.7 and later.
# ----------------------------------------------------------------
# Those who for any particular reason absolutely must score on
# Pentium can replace this module with one from 0.9.6 distribution.
# This "offer" shall be revoked the moment programming interface to
# this module is changed, in which case this paragraph should be
# removed.
# ----------------------------------------------------------------
#
$normal=0;
push(@INC,"perlasm","../../perlasm");
require "x86asm.pl";
&asm_init($ARGV[0],"sha1-586.pl",$ARGV[$#ARGV] eq "386");
$A="eax";
$B="ecx";
$C="ebx";
$D="edx";
$E="edi";
$T="esi";
$tmp1="ebp";
$off=9*4;
@K=(0x5a827999,0x6ed9eba1,0x8f1bbcdc,0xca62c1d6);
&sha1_block_data("sha1_block_asm_data_order");
&asm_finish();
sub Nn
{
local($p)=@_;
local(%n)=($A,$T,$B,$A,$C,$B,$D,$C,$E,$D,$T,$E);
return($n{$p});
}
sub Np
{
local($p)=@_;
local(%n)=($A,$T,$B,$A,$C,$B,$D,$C,$E,$D,$T,$E);
local(%n)=($A,$B,$B,$C,$C,$D,$D,$E,$E,$T,$T,$A);
return($n{$p});
}
sub Na
{
local($n)=@_;
return( (($n )&0x0f),
(($n+ 2)&0x0f),
(($n+ 8)&0x0f),
(($n+13)&0x0f),
(($n+ 1)&0x0f));
}
sub X_expand
{
local($in)=@_;
&comment("First, load the words onto the stack in network byte order");
for ($i=0; $i<16; $i+=2)
{
&mov($A,&DWP(($i+0)*4,$in,"",0));# unless $i == 0;
&mov($B,&DWP(($i+1)*4,$in,"
本文介绍了CTF挑战中的一个题目，该题目涉及到SHA1和MD5散列函数的碰撞利用。解题关键在于理解当传入SHA1函数的参数不是字符串时会返回null，通过构造`name[]=1&&password[]=5`的数组输入可以绕过条件。此外，还提到了MD5碰撞的概念，即不同输入可以产生相同的MD5值，这在攻击中可能被利用。建议遇到此类问题时，应及时搜索相关信息并调整payload。
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