如何将 uint256 变量转换为 int256 变量?

How do I convert a uint256 variable to an int256 one?

我试图通过在 return price; 正下方键入 return timeStamp; 来打印 uint timeStamp 代码:

pragma solidity ^0.6.7;

import "@chainlink/contracts/src/v0.6/interfaces/AggregatorV3Interface.sol";

contract PriceConsumerV3 {

AggregatorV3Interface internal priceFeed;

/**
 * Network: Kovan
 * Aggregator: BTC/USD
 * Address: 0x6135b13325bfC4B00278B4abC5e20bbce2D6580e
 */
constructor() public {
    priceFeed = AggregatorV3Interface(0x6135b13325bfC4B00278B4abC5e20bbce2D6580e);
}

/**
 * Returns the latest price
 */
function getThePrice() public view returns (int) {
    (
        uint80 roundID, 
        int price,
        uint startedAt,
        uint timeStamp,
        uint80 answeredInRound
    ) = priceFeed.latestRoundData();
    return price;
    return timeStamp;
}
}

当我在 Remix Compiler 上编译上面的代码时,它回答:

TypeError: Return argument type uint256 is not implicitly convertible to expected type (type of first return variable) int256. return timeStamp; ^-------^

我倾向于认为我只需要输入 int256 return timeStamp 或类似的东西而不是 return timeStamp; 但我想不通。

欢迎反馈。

您可以使用以下语法将 uint 类型转换为 int

return int(timeStamp);

注意:对于 2^255-1int 最大值)和 2^256-1uint 最大值)。但这很可能只是一个理论案例,因为时间戳预计不会有这么大的值。


请注意,此行无法访问,因为您已经 returning price 上一行。

(
    uint80 roundID, 
    int price,
    uint startedAt,
    uint timeStamp,
    uint80 answeredInRound
) = priceFeed.latestRoundData();
return price; // the `price` is returned, and the function doesn't execute after this line
return timeStamp; // this is ignored because of the early return on previous line

如果你想return多个值,你可以使用这个语法:

// note the multiple datatypes in the `returns` block
function getThePriceAndTimestamp() public view returns (int, uint) {
    (
        uint80 roundID, 
        int price,
        uint startedAt,
        uint timeStamp,
        uint80 answeredInRound
    ) = priceFeed.latestRoundData();
    return (price, timeStamp); // here returning multiple values
}