Question

Hopefully this isn’t a repeat, but there are 5000+ questions here with “not all code paths return a value”!

Quite simply, why does this method with a non-generic implementation compile just fine:

    public static async Task TimeoutAfter(this Task task, int millisecondsTimeout)
    {
        if (task == await Task.WhenAny(task, Task.Delay(millisecondsTimeout)))
            await task;
        else
            throw new TimeoutException();
    }

while this attempt to make the method generic generates a Return state missing / …not all code paths return a value warning / error?:

    public static async Task<T> TimeoutAfter<T>(this Task<T> task, int millisecondsTimeout)
    {
        if (task == await Task.WhenAny(task, Task.Delay(millisecondsTimeout)))
            await task;
        else
            throw new TimeoutException();
    }

Answer 1

The non-generic Task type is somewhat equivalent to an awaitable void method. Just like a void method, you can’t return anything from a method that has a return type of Task, which is why the first example compiles. The second example, though, expects a return value of the generic type and you aren’t providing one in the path where you await another call.

Quoting from the MSDN reference on the async keyword, specifically about return types.

You use Task if no meaningful value is returned when the method is
completed. That is, a call to the method returns a Task, but when the
Task is completed, any await expression that’s awaiting the Task
evaluates to void.