Passing yield as a function argument...

[Peter Otten]

Christopher Reimer wrote:

> Greetings,
> 
> I have two functions that I generalized to be nearly identical except
> for one line. One function has a yield statement, the other function
> appends to a queue.
> 
> If I rewrite the one line to be a function passed in as an argument --
> i.e., func(data) -- queue.append works fine. If I create and pass an
> inner function with the yield statement, nothing happens.
> 
> Is it possible to pass a yield statement in some form as a function
> argument?

No. A yield expression turns a function into a generator function, and it 
does so at compile time:

>>> def f(): pass
... 
>>> def g():
...     if 0: yield
... 
>>> inspect.isgeneratorfunction(f)
False
>>> inspect.isgeneratorfunction(g)
True

Perhaps you can go the other way and write the queue variant in terms of the 
generator, something like


def g(...):
    for ...:
        ...
        yield ...

def f(...):
    for x in g(...):
        queue.append(x)

or if your queue supports it just 

queue.extend(g(...))