2023 | CrewCTF | Web3

deception

#smart contract #fake source code

Description

Tate doesn't want you to know the truth. Find the secret.

nc deception.chal.crewc.tf 60002

Setup.sol

pragma solidity ^0.8.13;

import "./Deception.sol";

contract Setup {
    deception public immutable TARGET;

    constructor() payable {
        TARGET = new deception(); 
    }

    function isSolved() public view returns (bool) {
        return TARGET.solved();
    }
}

Deception.sol

// Contract that has to be displayed for challenge

// SPDX-License-Identifier: MIT
pragma solidity ^0.8.10;

contract deception{
    address private owner;
    bool public solved;

    constructor() {
    owner = msg.sender;
    solved = false;
    }

    modifier onlyOwner() {
    require(msg.sender==owner, "Only owner can access");
    _;
    }

    function changeOwner(address newOwner) onlyOwner public{
    owner = newOwner;
    }

    function password() onlyOwner public view returns(string memory){
        return "secret";
    }

    function solve(string memory secret) public {
        require(keccak256(abi.encodePacked(secret))==0x65462b0520ef7d3df61b9992ed3bea0c56ead753be7c8b3614e0ce01e4cac41b, "invalid");
        solved = true;
    }
}

Solution

• From the source code, if we are able to provide a secret whose keccak256 hash is equal to 0x65462b0520ef7d3df61b9992ed3bea0c56ead753be7c8b3614e0ce01e4cac41b, then the challenge can be solved. And, the keccak256 hash of "secret" is exactly what we want XD
• However, when I called the solve() function with the argument "secret", the transaction kept reverting :(
• I suddenly realized that the secret provided in the source code is not the actual value. So, I got the bytecode and attempted to extract the password from it

• It's not easy to get the secret even with decompiled bytecode

  function password() public payable { 
      require(msg.sender == _changeOwner, Error('Only owner can access'));
      v0 = _SafeExp(stor_1, stor_4);
      require(stor_2, Panic(18)); // division by zero
      if (76 - v0 % stor_2) {
          MEM[MEM[64] + 32] = v0 % stor_2;
          v1 = v2 = MEM[64] + 64;
      } else {
          require((stor_3 == stor_3 * (v0 % stor_2) / (v0 % stor_2)) | !(v0 % stor_2), Panic(17)); // arithmetic overflow or underflow
          require(v0 % stor_2, Panic(18)); // division by zero
          MEM[32 + MEM[64]] = stor_3 * (v0 % stor_2) / (v0 % stor_2);
          v3 = 0x18e(64 + MEM[64], 32, 30);
          v4 = _SafeAdd(0x616263, stor_3);
          v5 = _SafeSub(v4, stor_3);
          MEM[32 + MEM[64]] = v5;
          v6 = 0x18e(64 + MEM[64], 32, 30);
          v7 = v8 = 0;
          while (v7 < v3.length) {
              MEM[v7 + (32 + MEM[64])] = v3[v7];
              v7 += 32;
          }
          MEM[v3.length + (32 + MEM[64])] = 0;
          v9 = v10 = 0;
          while (v9 < v6.length) {
              MEM[v9 + (32 + MEM[64] + v3.length)] = v6[v9];
              v9 += 32;
          }
          MEM[v6.length + (32 + MEM[64] + v3.length)] = 0;
          v1 = v11 = v6.length + (32 + MEM[64] + v3.length);
      }
      v12 = new array[](v1 - MEM[64] - 32);
      v13 = v14 = 0;
      while (v13 < v1 - MEM[64] - 32) {
          MEM[v13 + v12.data] = MEM[v13 + (MEM[64] + 32)];
          v13 += 32;
      }
      MEM[v1 - MEM[64] - 32 + v12.data] = 0;
      return v12;
  }
  

• Why not just fork the chain and impersonate the owner to call the password() function?

  contract DeceptionTest is Test {
      Setup setup;
      deception target;
  
      function setUp() public {
          vm.createSelectFork(vm.envString("RPC_URL"));
          setup = Setup(vm.envAddress("INSTANCE_ADDR"));
          target = setup.TARGET();
      }
  
      function testSolve() public {
          // address private owner;
          bytes32 slotValue = vm.load(address(target), 0);
          vm.prank(address(uint160(uint256(slotValue))));
          console.log("%s", target.password());
      }
  }
  

• Call the solve() function to complete the challenge after getting the actual secret :D

Flag

crew{d0nt_tru5t_wh4t_y0u_s3e_4s5_50urc3!}

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最后更新: 2023年7月10日 10:55:12

Contributors: YanhuiJessica
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