在一段时间内执行一个函数,时间作为参数

我想用一种简单的异步方法:

async Task Loop(int timeInMs, int iterationTimeInMs)
{
    for (int current = 0; current < timeInMs; current += iterationTimeInMs)
    {
        await Task.Delay(iterationTimeInMs);
        var normalizedTime = current / (double)timeInMs;
        someValue = Lerp(from, to, normalizedTime);
    }
}

如果你的 Lerp 可能需要很长时间

async Task Loop(int timeInMs, int iterationTimeInMs)
{
    Task taskFromLastIteration = null;
    for (int current = 0; current < timeInMs; current += iterationTimeInMs);
    {            
        var delay = Task.Delay(iterationTimeInMs); // start delay now
        if (taskFromLastIteration != null)
            someValue = await taskFromLastIteration;
        await delay;
        var normalizedTime = current / (double)timeInMs;
        taskFromLastIteration = Task.Run(() => Lerp(from, to, normalizedTime));
    }
    if (taskFromLastIteration != null)
        someValue = await taskFromLastIteration;
}