LazyFund逻辑数值问题梳理

二测的逻辑是

    uint256 blockHeightDiff = block.number.sub(stream.lastRewardBlock);
    uint256 m = amount.mul(stream.kBlock).div(blockHeightDiff); //If underflow m might be 0
    uint256 noverk = blockHeightDiff.mul(ONE).div(stream.kBlock);
    uint256 mu = stream.unlockRatio.mul(ONE).div(1000);
    uint256 onesubmu = ONE.sub(mu);
    // uint256 s = m.mul(ONE.sub(XNum.bpow(onesubmu,noverk))).div(ONE).div(mu).mul(ONE);
    uint256 s = m.mul(ONE.sub(XNum.bpow(onesubmu, noverk))).div(mu);

    // update remaining and withdrawable balance
    stream.lastRewardBlock = block.number;
    stream.remaining = remaining.add(amount * 2).sub(s); // = remaining + amount + (amount - s)
    stream.withdrawable = withdrawable.add(s).sub(amount); // = withdrawable - amount + s

问题1：r和w其实不需要重新分配，就amount分配给s和剩余问题2: s是sum，应该给r，剩余给w 问题3: 指数操作其实有边界条件，需要考虑数值精度

建议 // Solve P3 uint256 blockHeightDiff = block.number.sub(stream.lastRewardBlock); uint256 m = amount.mul(ONE).div(blockHeightDiff); //If underflow m might be 0, peg true kBlock to 1, if bHD 0 then error uint256 noverk = blockHeightDiff.mul(ONE); // peg true kBlock to 1 so n over k always greater or equal 1 uint256 mu = stream.unlockRatio.mul(ONE).div(1000).div(stream.kBlock); // peg true mu to mu/kBlock uint256 onesubmu = ONE.sub(mu); // Enlarged due to mu divided by kBlock // uint256 s = m.mul(ONE.sub(XNum.bpow(onesubmu,noverk))).div(ONE).div(mu).mul(ONE); uint256 s = m.mul(ONE.sub(XNum.bpow(onesubmu, noverk))).div(mu).div(ONE);

    // update remaining and withdrawable balance
    stream.lastRewardBlock = block.number;
    // Solve P1/P2, keep amount distributed
    stream.remaining = remaining.add(s); // = remaining + s
    stream.withdrawable = withdrawable.add(amount).sub(s); // = w + amount - s

xdefilab / xdefi-governance-token

LazyFund逻辑数值问题梳理 #18