In python is there a way to check if a function is a “generator function” before calling it?

一曲冷凌霜 提交于 2019-11-28 17:23:42
>>> import inspect
>>> 
>>> def foo():
...   return 'foo'
... 
>>> def bar():
...   yield 'bar'
... 
>>> print inspect.isgeneratorfunction(foo)
False
>>> print inspect.isgeneratorfunction(bar)
True
  • New in Python version 2.6

Actually, I'm wondering just how useful such a hypothetical is_generator_function() would be really. Consider:

def foo():
    return 'foo'
def bar():
    yield 'bar'
def baz():
    return bar()
def quux(b):
    if b:
        return foo()
    else:
        return bar()

What should is_generator_function() return for baz and quux? baz() returns a generator but isn't one itself, and quux() might return a generator or might not.

>>> def foo():
...   return 'foo'
... 
>>> def bar():
...   yield 'bar'
... 
>>> import dis
>>> dis.dis(foo)
  2           0 LOAD_CONST               1 ('foo')
              3 RETURN_VALUE        
>>> dis.dis(bar)
  2           0 LOAD_CONST               1 ('bar')
              3 YIELD_VALUE         
              4 POP_TOP             
              5 LOAD_CONST               0 (None)
              8 RETURN_VALUE        
>>> 

As you see, the key difference is that the bytecode for bar will contain at least one YIELD_VALUE opcode. I recommend using the dis module (redirecting its output to a StringIO instance and checking its getvalue, of course) because this provides you a measure of robustness over bytecode changes -- the exact numeric values of the opcodes will change, but the disassembled symbolic value will stay pretty stable;-).

I've implemented a decorator that hooks on the decorated function returned/yielded value. Its basic goes:

import types
def output(notifier):
    def decorator(f):
        def wrapped(*args, **kwargs):
            r = f(*args, **kwargs)
            if type(r) is types.GeneratorType:
                for item in r:
                    # do something
                    yield item
            else:
                # do something
                return r
    return decorator

It works because the decorator function is unconditionnaly called: it is the return value that is tested.


EDIT: Following the comment by Robert Lujo, I ended up with something like:

def middleman(f):
    def return_result(r):
        return r
    def yield_result(r):
        for i in r:
            yield i
    def decorator(*a, **kwa):
        if inspect.isgeneratorfunction(f):
            return yield_result(f(*a, **kwa))
        else:
            return return_result(f(*a, **kwa))
    return decorator
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