如何將 byte32 數組的元素轉換為數字？

該數組包含一個 SHA256 雜湊。我想將 2 個十六進制值相加，然後將它們乘以 1.125（只保留整數）。結果數字將代表一個字母數字字元，它將是隨機的。

“0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ”（長度為36）

所以我想從這個字元串中選擇隨機字母，我擁有的是一個 SHA256 雜湊。

function GenerateSixDigitCode() private view returns (bytes32) {
   uint8[6] TwoHashDigit;                     //We add up 2 hex digit, than multiply it by 1.125 (32 to 36 conversion)

   //The CryptoZombies tutorial mentioned that this is not the best source of randomness, but in some cases this is sufficient.
   //This will be a ticket buying app just for learning purposes so it should be good enough 
   bytes32 seed = (keccak256(abi.encodePacked(
       block.timestamp + block.difficulty +
       ((uint256(keccak256(abi.encodePacked(block.coinbase)))) / (now)) +
       block.gaslimit + 
       ((uint256(keccak256(abi.encodePacked(msg.sender)))) / (now)) +
       block.number
   )));

   //The user should get a 6-digit code, like this: X4C9BA
   //I'm trying to generate the 6-digit code from sha256

   TwoHashDigit[0] = (seed[0] + seed[1]) * 1.125;
   TwoHashDigit[1] = (seed[2] + seed[3]) * 1.125;
   TwoHashDigit[2] = (seed[4] + seed[5]) * 1.125;
   TwoHashDigit[3] = (seed[6] + seed[7]) * 1.125;
   TwoHashDigit[4] = (seed[8] + seed[9]) * 1.125;
   TwoHashDigit[5] = (seed[10] + seed[11]) * 1.125;

   //TwoHashDigit should contain random numbers now, in the range of 0-36

   //Later I would map numbers to alphanumeric characters, using this string  alpha = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
   //For example alpha[10] == 'A'
   //alpha[20] == 'K'
}

我得到的錯誤：

contracts/ticketpurchaser.sol:43:27: TypeError: Operator * not compatible with types bytes1 and rational_const 9 / 8
       TwoHashDigit[5] = (seed[10] + seed[11]) * 1.125;

更改每次出現的情況：

(seed[x] + seed[y]) * 1.125

對此：

uint8((uint16(seed[x]) + uint16(seed[y])) * 9 / 8)

既然你有很多，你也可以在一個函式中實現它：

function combine(bytes1 x, bytes1 y) private pure returns (uint8) {
   return uint8((uint16(x) + uint16(y)) * 9 / 8);
}

引用自：https://ethereum.stackexchange.com/questions/80672