How does Task<int> become an int?

Does an implicit conversion occur between Task<> and int?

Nope. This is just part of how async/await works.

Any method declared as async has to have a return type of:

• void (avoid if possible)
• Task (no result beyond notification of completion/failure)
• Task<T> (for a logical result of type T in an async manner)

The compiler does all the appropriate wrapping. The point is that you're asynchronously returning urlContents.Length - you can't make the method just return int, as the actual method will return when it hits the first await expression which hasn't already completed. So instead, it returns a Task<int> which will complete when the async method itself completes.

Note that await does the opposite - it unwraps a Task<T> to a T value, which is how this line works:

string urlContents = await getStringTask;

... but of course it unwraps it asynchronously, whereas just using Result would block until the task had completed. (await can unwrap other types which implement the awaitable pattern, but Task<T> is the one you're likely to use most often.)

This dual wrapping/unwrapping is what allows async to be so composable. For example, I could write another async method which calls yours and doubles the result:

public async Task<int> AccessTheWebAndDoubleAsync()
{
    var task = AccessTheWebAsync();
    int result = await task;
    return result * 2;
}

(Or simply return await AccessTheWebAsync() * 2; of course.)

No requires converting the Task to int. Simply Use The Task Result.

int taskResult = AccessTheWebAndDouble().Result;

public async Task<int> AccessTheWebAndDouble()
{
    int task = AccessTheWeb();
    return task;
}

It will return the value if available otherwise it return 0.