Answer: Why is this code broken?

Apr 07 2022

AnswerWhy is this code broken?

---

I asked why this code is broken, and now is the time to dig into this. The issue is in this block of code. Take a look at that for a moment, if you please:

tasks.Add(Task.Factory.StartNew(async () => { using var data = File.OpenRead(file); var url = "/file?=" + Uri.EscapeDataString(Path.GetFileName(file)); await httpClient.PostAsync(url, new StreamContent(data)); }));

The code is trying to gather the async upload of all the files, and then it will await them. This code compile and runs successfully, but it will not do what you expect it to do. Let’s break it up a bit to understand what is going on:

var task = Task.Factory.StartNew(async () => { // redacted }) tasks.Add(task);

We are passing the variable task to the list of tasks we have. We just extract a variable, nothing much going on here. Let’s explore further, what is the type of task? We know it must be a subtype of Task, since that is what the tasks collection will take. It turns out that this isn’t that simple:

Task<Task> task = Task.Factory.StartNew(async () => { // redacted }) tasks.Add(task);

What is that, Task<Task> thingie? Well, let’s look at the signature of Task.Factory.StartNew(), shall we?

public Task<TResult> StartNew<TResult>(Func<TResult> function);

Just from the signature, we can tell what is going on. StartNew() accepts a function that returns a value and will then return a task for the eventual result of this function. However, the function we are actually passing to the StartNew() doesn’t produce a value. Except that it does…

Let’s explore that thing for a bit:

var func = async () => { };

What is the type of func in this case?

Func<Task> func = async() => {};

The idea is that when the compiler sees the async keyword, it transforms the function to one that returns a Task. Combining both those features together means that our original code actually registers the start of the asynchronous process to happen and will return as soon as it got started. Basically, we’ll only wait for the actual opening of the file, not for the actual network work that has to happen here.

The right way to express what we want here is:

Task.Run(async () => {});

The signature for this is:

public static Task Run<TResult>(Func<Task> function);

You can see here that we get a function that returns a task, but we aren’t actually wrapping that in another Task instance. The task that will be returned will be completed once the full work has been completed.

It is an interesting pitfall in the API, and can be quite hard to figure out exactly what is going on. Mostly because there are several different things happening all at once.