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问将字节转换为uint256[]
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Ethereum用户

提问于 2018-12-12 15:33:46

回答 1查看 1K关注 0票数 0

我正在编写ERC777TokensRecipient的实现。所以这意味着一个函数

function tokensReceived(address operator, address from, address to, uint256 amount, bytes data, bytes operatorData) external;

data字段包含14个uint256变量

0x000000000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000020000000000000000000000000000000000000000000000000000000000000003000000000000000000000000000000000000000000000000000000000000000400000000000000000000000000000000000000000000000000000000000000050000000000000000000000000000000000000000000000000000000000000006000000000000000000000000000000000000000000000000000000000000000700000000000000000000000000000000000000000000000000000000000000080000000000000000000000000000000000000000000000000000000000000009000000000000000000000000000000000000000000000000000000000000000a000000000000000000000000000000000000000000000000000000000000000b000000000000000000000000000000000000000000000000000000000000000c000000000000000000000000000000000000000000000000000000000000000d000000000000000000000000000000000000000000000000000000000000000e

我如何才能访问这些变量？

我知道我可以让他们这样：

function tokensReceived(address _operator, address _from, address _to, uint256 _amount, bytes _data, bytes _operatorData) external {
    uint arg1;
    uint arg2;
    uint arg3;
    uint arg4;
    uint arg5;
    uint arg6;
    uint arg7;
    uint arg8;
    uint arg9;
    uint arg10;
    uint arg11;
    uint arg12;
    uint arg13;
    uint arg14;

    bytes memory tmp = _data;
    assembly {
        arg1  := mload(add(0x020, tmp))
        arg2  := mload(add(0x040, tmp))
        arg3  := mload(add(0x060, tmp))
        arg4  := mload(add(0x080, tmp))
        arg5  := mload(add(0x0A0, tmp))
        arg6  := mload(add(0x0C0, tmp))
        arg7  := mload(add(0x0E0, tmp))
        arg8  := mload(add(0x100, tmp))
        arg9  := mload(add(0x120, tmp))
        arg10 := mload(add(0x140, tmp))
        arg11 := mload(add(0x160, tmp))
        arg12 := mload(add(0x180, tmp))
        arg13 := mload(add(0x1A0, tmp))
        arg14 := mload(add(0x1C0, tmp))
    }
}

但是，在遇到

CompilerError: Stack too deep, try removing local variables.

根据编译时出错:堆栈太深的说法，解决办法是把当地的uints放到一个数组中。但我不知道怎么做。

有人知道我怎么写这样的函数吗？

function convertToArrayOfLength14(bytes memory data) internal pure returns (uint[]) {
    // TODO
}

solidity

assembly

回答 1

Ethereum用户

回答已采纳

发布于 2018-12-12 15:46:14

你可以在这样的循环中这样做：

function convertToArrayOfLength14(bytes memory data) public pure returns (uint256[] memory output)
{
    output = new uint256[](14);
    for (uint256 i=32; i<=output.length*32; i+=32)
    {
        assembly { mstore(add(output, i), mload(add(data, i))) }
    }
}

票数 2

页面原文内容由Ethereum提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://ethereum.stackexchange.com/questions/63994

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