用于字节操作的 Solidity 代码无法使用 Solidity 0.8.0 的安全帽编译器进行编译

Question

我正在使用0.8.0 编译器从用Sol 0.5.0 编写的OpenSea 项目编译代码，但出现错误：

  ParserError: Expected primary expression.
--> contracts/Strings.sol:53:25:
 |
53 |             bstr[k--] = byte(uint8(48 + _i % 10));
 |                         ^^^^


Error HH600: Compilation failed

原始代码位于：https://github.com/ProjectOpenSea/opensea-creatures/blob/master/contracts/Strings.sol，它使用Sol 0.5.0，大概是用松露编译的。我正在尝试使用Hardhat 和0.8.0。代码转载如下：

pragma solidity ^0.8.0;

library Strings {
  // via https://github.com/oraclize/ethereum-api/blob/master/oraclizeAPI_0.5.sol
  function strConcat(string memory _a, string memory _b, string memory _c, string memory _d, string memory _e) internal pure returns (string memory) {
      bytes memory _ba = bytes(_a);
      bytes memory _bb = bytes(_b);
      bytes memory _bc = bytes(_c);
      bytes memory _bd = bytes(_d);
      bytes memory _be = bytes(_e);
      string memory abcde = new string(_ba.length + _bb.length + _bc.length + _bd.length + _be.length);
      bytes memory babcde = bytes(abcde);
      uint k = 0;
      for (uint i = 0; i < _ba.length; i++) babcde[k++] = _ba[i];
      for (uint i = 0; i < _bb.length; i++) babcde[k++] = _bb[i];
      for (uint i = 0; i < _bc.length; i++) babcde[k++] = _bc[i];
      for (uint i = 0; i < _bd.length; i++) babcde[k++] = _bd[i];
      for (uint i = 0; i < _be.length; i++) babcde[k++] = _be[i];
      return string(babcde);
    }

    function strConcat(string memory _a, string memory _b, string memory _c, string memory _d) internal pure returns (string memory) {
        return strConcat(_a, _b, _c, _d, "");
    }

    function strConcat(string memory _a, string memory _b, string memory _c) internal pure returns (string memory) {
        return strConcat(_a, _b, _c, "", "");
    }

    function strConcat(string memory _a, string memory _b) internal pure returns (string memory) {
        return strConcat(_a, _b, "", "", "");
    }

    function uint2str(uint _i) internal pure returns (string memory _uintAsString) {
        if (_i == 0) {
            return "0";
        }
        uint j = _i;
        uint len;
        while (j != 0) {
            len++;
            j /= 10;
        }
        bytes memory bstr = new bytes(len);
        uint k = len - 1;
        while (_i != 0) {
            bstr[k--] = byte(uint8(48 + _i % 10));
            _i /= 10;
        }
        return string(bstr);
    }
}

请注意，我在顶部更改了 pragma。对我来说一切都很好，所以我不确定问题出在哪里，除了它在这条线上： bstr[k--] = byte(uint8(48 + _i % 10));

Answer 1

使用bytes1 而不是byte。

---

类型byte 已被删除。它是bytes1 的别名。

来源：https://docs.soliditylang.org/en/v0.8.3/080-breaking-changes.html#silent-changes-of-the-semantics