Simpy 3: Resources.Resource.request()/.release() WITHOUT 'with...as:'

Question

I'm trying to add SimPy simulation to a project I'm working on and I have some confusion about version 3's release/request.

I was able to implement resources using a 'with'-block without trouble but in my situation I want to request/release a resource without using a 'with'-block.

However, I cannot find an example of this using SimPy 3. I read the documentation/source regarding resources but still can't get it quite right. Could someone explain how to properly:

...
Request a Resource with the method: 'request()'
...
Release that Resource with the method: 'release()'
...

Thanks, and sorry for the bother.

PS: I'm intending to use Resources.resource

Answer 1

If you want to use a resource without a with block (and you know you won’t get interrupted), its just:

req = resource.request()
yield req
# do stuff
resource.release(req)

Answer 2

Using with on an object calls __enter__ when you enter the with block, and __exit__ when you leave. So when you do

res = resource.Resource()
with res.request() as req:
  # stuff

You're really calling __enter__ on a Request object, doing #stuff then calling __exit__:

class Request(base.Put):
    def __exit__(self, exc_type, value, traceback):
        super(Request, self).__exit__(exc_type, value, traceback)
        self.resource.release(self)

class Put(Event):  # base.Put
    def __enter__(self):
        return self

    def __exit__(self, exc_type, exc_value, traceback):
        # If the request has been interrupted, remove it from the queue:
        if not self.triggered:
            self.resource.put_queue.remove(self)

So, the with block is equivalent to this:

res = resource.Resource(...)
req = res.request()
#stuff
if not req.triggered:
   res.put_queue.remove(req)
   res.release(req)

However, the with block is also making sure that the cleanup code is called no matter what exceptions are thrown during #stuff. You'll lose that with the above code.