下面是一个程序的核心算法，请大家说一下大概是 什么意思？谢谢
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004021CC > \8D8D D4F9FFFF lea ecx,dword ptr ss:[ebp-62C]
004021D2 . 8D95 D4FDFFFF lea edx,dword ptr ss:[ebp-22C]
004021D8 > 8A01 mov al,byte ptr ds:[ecx]
004021DA . 41 inc ecx
004021DB . 8802 mov byte ptr ds:[edx],al
004021DD . 42 inc edx
004021DE . 84C0 test al,al
004021E0 .^ 75 F6 jnz short 复件_kid.004021D8
004021E2 . 8D95 D4F9FFFF lea edx,dword ptr ss:[ebp-62C]
004021E8 . 52 push edx
004021E9 . E8 C2F9FFFF call 复件_kid.00401BB0
004021EE . 8B4D D0 mov ecx,dword ptr ss:[ebp-30]
004021F1 . 83C4 04 add esp,4
004021F4 . 8D95 D4F5FFFF lea edx,dword ptr ss:[ebp-A2C]
004021FA > 8A01 mov al,byte ptr ds:[ecx]
004021FC . 41 inc ecx
004021FD . 8802 mov byte ptr ds:[edx],al
004021FF . 42 inc edx
00402200 . 84C0 test al,al
00402202 .^ 75 F6 jnz short 复件_kid.004021FA
00402204 . 8DB5 D4F5FFFF lea esi,dword ptr ss:[ebp-A2C]
0040220A . 8D85 D4F9FFFF lea eax,dword ptr ss:[ebp-62C]
00402210 > 8A10 mov dl,byte ptr ds:[eax]
00402212 . 8ACA mov cl,dl
00402214 . 3A16 cmp dl,byte ptr ds:[esi]
00402216 . 75 1C jnz short 复件_kid.00402234
00402218 . 84C9 test cl,cl
0040221A . 74 14 je short 复件_kid.00402230
0040221C . 8A50 01 mov dl,byte ptr ds:[eax+1]
0040221F . 8ACA mov cl,dl
00402221 . 3A56 01 cmp dl,byte ptr ds:[esi+1]
00402224 . 75 0E jnz short 复件_kid.00402234
00402226 . 83C0 02 add eax,2
00402229 . 83C6 02 add esi,2
0040222C . 84C9 test cl,cl
0040222E .^ 75 E0 jnz short 复件_kid.00402210
00402230 > 33C0 xor eax,eax
00402232 . EB 05 jmp short 复件_kid.00402239
00402234 > 1BC0 sbb eax,eax
00402236 . 83D8 FF sbb eax,-1
00402239 > 85C0 test eax,eax
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[招生]科锐逆向工程师培训(2024年11月15日实地，远程教学同时开班, 第51期)
