如何在solidity中将uint转换为字符串？

Question

在 Solidity 中，有没有办法可以将我的 int 转换为字符串？

例子：

pragma solidity ^0.4.4;

contract someContract {

    uint i;

    function test() pure returns (string) {

      return "Here and Now is Happiness!";

    }

    function love() pure returns(string) {

        i = i +1;

        return "I love " + functionname(i) + " persons" ;
    }



}

什么是函数名？谢谢！

Answer 1

Solidity 0.8.0 更新：

来自https://github.com/provable-things/ethereum-api/blob/master/provableAPI_0.6.sol的uint2str()函数现在已经过时，无法工作，但这里是更新的代码，它使用了solidity 0.8.0 ：（在上一版本中有一个溢出错误，但solidity <0.8.0 忽​​略了它，因为它不影响答案，但现在抛出错误） byte也更改为bytes1和+、-、* 和就像他们从 SafeMath 库中那样工作。

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;
        while (_i != 0) {
            k = k-1;
            uint8 temp = (48 + uint8(_i - _i / 10 * 10));
            bytes1 b1 = bytes1(temp);
            bstr[k] = b1;
            _i /= 10;
        }
        return string(bstr);
    }

Answer 2

这里的两个帖子给出了回应：

https://ethereum.stackexchange.com/questions/10811/solidity-concatenate-uint-into-a-string

https://ethereum.stackexchange.com/questions/10932/how-to-convert-string-to-int

function uintToString(uint v) constant returns (string str) {
        uint maxlength = 100;
        bytes memory reversed = new bytes(maxlength);
        uint i = 0;
        while (v != 0) {
            uint remainder = v % 10;
            v = v / 10;
            reversed[i++] = byte(48 + remainder);
        }
        bytes memory s = new bytes(i + 1);
        for (uint j = 0; j <= i; j++) {
            s[j] = reversed[i - j];
        }
        str = string(s);
    }

问候

Answer 3

在对已接受答案的评论中建议的可证明事物代码对我"uintToStr": Avoid assigning to function parameters. [security/no-assign-params] ，但我的"uintToStr": Avoid assigning to function parameters. [security/no-assign-params]发出了一个警告，即： "uintToStr": Avoid assigning to function parameters. [security/no-assign-params] "uintToStr": Avoid assigning to function parameters. [security/no-assign-params] 。 下面稍微更改了原始代码以更正此问题（将参数_i重新分配给另一个名为number变量）：

    /// @notice converts number to string
    /// @dev source: https://github.com/provable-things/ethereum-api/blob/master/oraclizeAPI_0.5.sol#L1045
    /// @param _i integer to convert
    /// @return _uintAsString
    function uintToStr(uint _i) internal pure returns (string memory _uintAsString) {
        uint number = _i;
        if (number == 0) {
            return "0";
        }
        uint j = number;
        uint len;
        while (j != 0) {
            len++;
            j /= 10;
        }
        bytes memory bstr = new bytes(len);
        uint k = len - 1;
        while (number != 0) {
            bstr[k--] = byte(uint8(48 + number % 10));
            number /= 10;
        }
        return string(bstr);
    }

Answer 4

Concrete_Buddhas 答案在 Solidity 0.8.0 中不起作用。 这是修订版：

 function uint2str(
  uint256 _i
)
  internal
  pure
  returns (string memory str)
{
  if (_i == 0)
  {
    return "0";
  }
  uint256 j = _i;
  uint256 length;
  while (j != 0)
  {
    length++;
    j /= 10;
  }
  bytes memory bstr = new bytes(length);
  uint256 k = length;
  j = _i;
  while (j != 0)
  {
    bstr[--k] = bytes1(uint8(48 + j % 10));
    j /= 10;
  }
  str = string(bstr);
}

Answer 5

如果您需要选择性地转换为科学记数法，例如为了更紧凑的数字表示，这里有一个用于此目的的修改版本：

function uintToString(uint v, bool scientific) public pure returns (string memory str) {

    if (v == 0) {
        return "0";
    }

    uint maxlength = 100;
    bytes memory reversed = new bytes(maxlength);
    uint i = 0;
    
    while (v != 0) {
        uint remainder = v % 10;
        v = v / 10;
        reversed[i++] = byte(uint8(48 + remainder));
    }

    uint zeros = 0;
    if (scientific) {
        for (uint k = 0; k < i; k++) {
            if (reversed[k] == '0') {
                zeros++;
            } else {
                break;
            }
        }
    }

    uint len = i - (zeros > 2 ? zeros : 0);
    bytes memory s = new bytes(len);
    for (uint j = 0; j < len; j++) {
        s[j] = reversed[i - j - 1];
    }

    str = string(s);

    if (scientific && zeros > 2) {
        str = string(abi.encodePacked(s, "e", uintToString(zeros, false)));
    }
}

一些单元测试：

function testUintToString() public {

    Assert.equal(Utils.uintToString(0, true), '0', '0');
    Assert.equal(Utils.uintToString(1, true), '1', '1');
    Assert.equal(Utils.uintToString(123, true), '123', '123');
    Assert.equal(Utils.uintToString(107680546035, true), '107680546035', '107680546035');
    Assert.equal(Utils.uintToString(1e9, true), '1e9', '1e9');
    Assert.equal(Utils.uintToString(1 ether, true), '1e18', '1 ether');
    Assert.equal(Utils.uintToString(550e8, true), '55e9', '55e9');
}

上面的代码片段与solidity 0.6.0兼容。

Answer 6

solidity ^0.8.0

import "@openzeppelin/contracts/utils/Strings.sol";

Strings.toString(myUINT)

为我工作。

https://github.com/OpenZeppelin/openzeppelin-contracts/blob/master/contracts/utils/Strings.sol#L15-L35

Answer 7

虽然公认的答案似乎是正确的，但在大量数据上效率很低。 这是我的做法：

function itoa32 (uint x) private pure returns (uint y) {
    unchecked {
        require (x < 1e32);
        y = 0x3030303030303030303030303030303030303030303030303030303030303030;
        y += x % 10; x /= 10;
        y += x % 10 << 8; x /= 10;
        y += x % 10 << 16; x /= 10;
        y += x % 10 << 24; x /= 10;
        y += x % 10 << 32; x /= 10;
        y += x % 10 << 40; x /= 10;
        y += x % 10 << 48; x /= 10;
        y += x % 10 << 56; x /= 10;
        y += x % 10 << 64; x /= 10;
        y += x % 10 << 72; x /= 10;
        y += x % 10 << 80; x /= 10;
        y += x % 10 << 88; x /= 10;
        y += x % 10 << 96; x /= 10;
        y += x % 10 << 104; x /= 10;
        y += x % 10 << 112; x /= 10;
        y += x % 10 << 120; x /= 10;
        y += x % 10 << 128; x /= 10;
        y += x % 10 << 136; x /= 10;
        y += x % 10 << 144; x /= 10;
        y += x % 10 << 152; x /= 10;
        y += x % 10 << 160; x /= 10;
        y += x % 10 << 168; x /= 10;
        y += x % 10 << 176; x /= 10;
        y += x % 10 << 184; x /= 10;
        y += x % 10 << 192; x /= 10;
        y += x % 10 << 200; x /= 10;
        y += x % 10 << 208; x /= 10;
        y += x % 10 << 216; x /= 10;
        y += x % 10 << 224; x /= 10;
        y += x % 10 << 232; x /= 10;
        y += x % 10 << 240; x /= 10;
        y += x % 10 << 248;
    }
}

function itoa (uint x) internal pure returns (string memory s) {
    unchecked {
        if (x == 0) return "0";
        else {
            uint c1 = itoa32 (x % 1e32);
            x /= 1e32;
            if (x == 0) s = string (abi.encode (c1));
            else {
                uint c2 = itoa32 (x % 1e32);
                x /= 1e32;
                if (x == 0) {
                    s = string (abi.encode (c2, c1));
                    c1 = c2;
                } else {
                    uint c3 = itoa32 (x);
                    s = string (abi.encode (c3, c2, c1));
                    c1 = c3;
                }
            }
            uint z = 0;
            if (c1 >> 128 == 0x30303030303030303030303030303030) { c1 <<= 128; z += 16; }
            if (c1 >> 192 == 0x3030303030303030) { c1 <<= 64; z += 8; }
            if (c1 >> 224 == 0x30303030) { c1 <<= 32; z += 4; }
            if (c1 >> 240 == 0x3030) { c1 <<= 16; z += 2; }
            if (c1 >> 248 == 0x30) { z += 1; }
            assembly {
                let l := mload (s)
                s := add (s, z)
                mstore (s, sub (l, z))
            }
        }
    }
}

解释

itoa32 function 将低于 10^32 的数字转换为 32 位数字，必要时用零填充。

itoa function 最多调用itoa32 3 次以转换任意 256 位数字，然后连接结果并删除前导零。 它使用二进制搜索来找出要删除的前导零的确切数量，并就地从字符串中删除前导零。

Answer 8

如果您的 uint 来自最初将字符串编码为 uint

string(abi.encode(myUint))