前言

兜兜转转终于回到这里了。之前复现了羊城杯较简单点的三题，到这一题vm卡住了，刷了好几道，终于搞定了。

题目分析

第一次遇到Mac os的题目，幸好还可以做。

opcode 和 指令操作

密文：

opcode先提取出来，得到：

opcode = [26, 0, 3, 25, 1, 1, 13, 2, 7, 24, 1, 2, 1, 0, 3, 26, 0, 3, 25,
          1, 2, 13, 2, 6, 24, 1, 2, 1, 0, 4, 26, 0, 3, 25, 1, 3, 13, 2, 5,
          24, 1, 2, 1, 0, 5, 26, 0, 3, 25, 1, 4, 13, 2, 4, 24, 1, 2, 1, 0, 6]

注意这里是，后面的八位拼接上去会把前面的最后一个给覆盖掉。同时数据在内存中是小端序存储，因此还需要逆序一下。

查个重。得到：

[26, 0, 3, 25, 1, 13, 2, 7, 24, 6, 4, 5]

一共26条指令，真的多，但是因为我们opcode很少，可以只还原一部分就OK了。

还原汇编

opcode = [26, 0, 3, 25, 1, 1, 13, 2, 7, 24, 1, 2, 1, 0, 3, 26, 0, 3, 25,
          1, 2, 13, 2, 6, 24, 1, 2, 1, 0, 4, 26, 0, 3, 25, 1, 3, 13, 2, 5,
          24, 1, 2, 1, 0, 5, 26, 0, 3, 25, 1, 4, 13, 2, 4, 24, 1, 2, 1, 0, 6]
# [26, 0, 3, 25, 1, 13, 2, 7, 24, 6, 4, 5]

i = 0
while opcode[i]:
    match opcode[i]:
        case 0:
            print(f"{i} swap reg%d reg%d" % (opcode[i+1],opcode[i+2]))
        case 1:
            print(f"{i} xor reg%d reg%d" % (opcode[i+1],opcode[i+2]))
        case 2:
            print(f"{i} add reg%d %d" % (opcode[i+1],opcode[i+2]))
        case 3:
            print(f"{i} add reg%d reg%d" % (opcode[i+1],opcode[i+2]))
        case 4:
            print(f"{i} sub reg%d %d" % (opcode[i+1],opcode[i+2]))
        case 5:
            print(f"{i} sub reg%d reg%d" % (opcode[i+1],opcode[i+2]))
        case 6:
            print(f"{i} mul reg%d %d" % (opcode[i+1],opcode[i+2]))
        case 7:
            print(f"{i} mul reg%d reg%d" % (opcode[i+1],opcode[i+2]))
        case 13:
            print(f"{i} mov reg%d reg0<<%d" % (opcode[i+1],opcode[i+2]))
        case 24:
            print(f"{i} reg0 = reg2 | reg1")
        case 25:
            print(f"{i} mov reg%d reg0>>%d" % (opcode[i+1],opcode[i+2]))
        case 26:
            print(f"{i} mov reg%d %d" % (opcode[i+1],opcode[i+2]))
    i += 3

运行后打印的结果1：

0 mov reg0 3
3 mov reg1 reg0>>1
6 mov reg2 reg0<<7
9 reg0 = reg2 | reg1
12 xor reg0 reg3
15 mov reg0 3
18 mov reg1 reg0>>2
21 mov reg2 reg0<<6
24 reg0 = reg2 | reg1
27 xor reg0 reg4
30 mov reg0 3
33 mov reg1 reg0>>3
36 mov reg2 reg0<<5
39 reg0 = reg2 | reg1
42 xor reg0 reg5
45 mov reg0 3
48 mov reg1 reg0>>4
51 mov reg2 reg0<<4
54 reg0 = reg2 | reg1
57 xor reg0 reg6

这里打印的指令蛮少的，而且也没有和密文比较的地方，推测是单字符加密，解密的话就逆向就好了。

加密逻辑写出来大概就是

	s = (s >> 1 |  s << 7) & 0xFF
    s = s ^ reg3
    s = (s >> 2 | s << 6) & 0xFF
    s = s ^ reg4
    s = (s >> 3 | s << 5) & 0xFF
    s = s ^ reg5
    s = (s >> 4 | s << 4) & 0xFF
    s = s ^ reg6

通过内存可以发现，reg3 到 reg6 就是

exp

enc = [0xDF, 0xD5, 0xF1, 0xD1, 0xFF, 0xDB, 0xA1, 0xA5, 0x89, 0xBD, 0xE9, 0x95, 0xB3, 0x9D, 0xE9, 0xB3, 0x85, 0x99, 0x87,
       0xBF, 0xE9, 0xB1, 0x89, 0xE9, 0x91, 0x89, 0x89, 0x8F, 0xAD]
key = 0xBEEDBEEF.to_bytes(4, 'little')


def decode(s):
    s = s ^ key[3]
    s = (s << 4 | s >> 4) & 0xFF
    s = s ^ key[2]
    s = (s << 3 | s >> 5) & 0xFF
    s = s ^ key[1]
    s = (s << 2 | s >> 6) & 0xFF
    s = s ^ key[0]
    s = (s << 1 | s >> 7) & 0xFF
    return s


flag = []
for i in enc:
    i = ((i >> 3) | (i << 5) & 0xFF)
    flag.append(decode(i))
print(bytes(flag))
# b'DASCTF{you_are_right_so_cool}'

后言

前面练的。
VM基础知识和EzMachine复现
HGAME 2023 week4vm复现
GWCTF 2019babyvm 复现
NCTF 2018wcyvm 复现