用于字节操作的Solidity代码无法使用Solidity为0.8.0的hardhat编译器进行编译

Question

我正在使用0.8.0编译器编译用Sol 0.5.0编写的OpenSea项目中的代码，并收到错误：

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，并且可能是用块菌编译的。我正在尝试使用Hardhat和0.8.0。代码复制如下：

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);
}
}

注意，我在顶部更改了pragma。在我看来一切都很好，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-语义的变化