Privacy

题目源码

// SPDX-License-Identifier: MIT
pragma solidity ^0.6.0;

contract Privacy {

  bool public locked = true;
  uint256 public ID = block.timestamp;
  uint8 private flattening = 10;
  uint8 private denomination = 255;
  uint16 private awkwardness = uint16(now);
  bytes32[3] private data;

  constructor(bytes32[3] memory _data) public {
    data = _data;
  }

  function unlock(bytes16 _key) public {
    require(_key == bytes16(data[2]));
    locked = false;
  }

  /*
    A bunch of super advanced solidity algorithms...

      ,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`
      .,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,
      *.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^         ,---/V\
      `*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.    ~|__(o.o)
      ^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'^`*.,*'  UU  UU
  */
}

题目要求

将锁解开，即将locked设置为true

题目分析

想要解锁，需要调用unlock方法，并正确传入_key。也就是要想办法获取到bytes16(data[2])的值。由于data是private的，所以要通过直接读取storage来获取data[2]的值

根据变量定义我们可以看到Storage布局如下

---------------------------------------------------------
| unused (31)                         | locked (1)       | <- slot 0
---------------------------------------------------------
|                      ID (32)                           | <- slot 1
---------------------------------------------------------
| awkwardness (16) | denomination(8) | flattening(8)     | <- slot 2
---------------------------------------------------------
                    data[0]                              | <- slot 3
---------------------------------------------------------
                    data[1]                              | <- slot 4
---------------------------------------------------------
                    data[2]                              | <- slot 5
---------------------------------------------------------

我们要读取data[2]的值，通过下面方式获取

// instance即合约地址。5是指第5个slot
await web3.eth.getStorageAt(instance,5) => 0x1a231b273bb6cd8aeb60e7568697a851d822cc66662ead8548194923946295ce
// unlock方法是传bytes16类型,所以截取前16位即可
await contract.unlock("0x1a231b273bb6cd8aeb60e7568697a851")

补充

如果这里data是一个动态数组，则slot的布局结构如下

---------------------------------------------------------
| unused (31)                         | locked (1)      | <- slot 0
---------------------------------------------------------
|                      ID (32)                          | <- slot 1
---------------------------------------------------------
| awkwardness (16) | denomination(8) | flattening(8)    | <- slot 2
---------------------------------------------------------
|                    data.length(32)                    | <- slot 3
---------------------------------------------------------
|                    data[0]                            | <- slot keccak256(3)
---------------------------------------------------------
|                    data[1]                            | <- slot keccak256(3)+1
---------------------------------------------------------
|                    data[2]                            | <- slot keccak256(3)+2
---------------------------------------------------------