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I know that if a "yield" statement is present in a function it will be treated as a generator. But how does python interpreter works in that case when the function(generator) called for the first time.I mean when the code is interpreted i think first "gen" will get bind to a function object. Why it doesn't execute the statements before yield unlike normal functions or classes when called for first time. How does it put a hold on the execution of the print function before yield

>>> def gen():
...    print("In gen")
...    yield 1
...    yield 2
... 
>>> type(gen)
<type 'function'>
>>> a = gen
>>> type(a)
<type 'function'>
>>> 
>>> b = a()
>>> type(b)
<type 'generator'>
>>> b.next()
In gen
1
>>> b.next()
2
>>>
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  • 2
    Because Python generators are, by definition, lazy. They only start yielding output once you start consuming them (i.e. iterate over them). Commented Sep 14, 2014 at 16:51
  • I mean before going to yield , why it doesn't execute the statements before yield. How does the interpreter know about it. Commented Sep 14, 2014 at 16:53
  • 1
    In addition to @unutbu's answer: Because Python treats a function body with a yield statement differently. Once you put a yield in there, it knows you'll be defining a generator, not a regular function. Try putting yield and return (with an argument) in the same function, Python won't let you compile that code, even if you never use that function. Commented Sep 14, 2014 at 17:00
  • @Lukas : Cool, thanks. A little more explanation on def executable will help. Commented Sep 14, 2014 at 17:02
  • See The yield statement: "Using a yield statement in a function definition is sufficient to cause that definition to create a generator function instead of a normal function." How exactly the Python bytecode compilation process works is highly involved and shouldn't have to concern you. Commented Sep 14, 2014 at 17:19

1 Answer 1

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When Python parses the def statement, it decides if the code is defining a generator or a function. If the code contains a yield expression, then it is a generator. So when the generator function a is called, it returns a generator object, b.

The code inside the def statement is not executed until b.next() is called.

Consider these two functions:

def gen():
     print('In gen')
     yield 1
     yield 2


def func():
     print('In gen')

Although gen and func are both functions, Python knows that gen is a generator:

In [96]: import inspect

In [101]: inspect.isgeneratorfunction(gen)
Out[101]: True

In [102]: inspect.isgeneratorfunction(func)
Out[102]: False

If you look inside the inspect module you see

def isgeneratorfunction(object):
    """Return true if the object is a user-defined generator function.

    Generator function objects provides same attributes as functions.

    See help(isfunction) for attributes listing."""
    return bool((isfunction(object) or ismethod(object)) and
                object.func_code.co_flags & CO_GENERATOR)

So apparently Python checks this first before executing the code inside the function.

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7 Comments

How does the def statement does it without reaching the control to the block inside the function. How does it find it?
I don't how how def works, so a little more explanation will help. I thought it just creates a function object and bind a name to it.
All of the code gets compiled. The compiler inspects the contents of the function definition, finds a yield statement, and determines this object is, in fact, a generator.
@kingsdeb You can read some documentation here.
@kingdeb: Although dis.dis(gen) shows no indication that gen is a generator function, gen.func_code.co_flags has a flag indicating that gen is a generator. So the def statement is creating a function object that knows its a generator. Does this answer your question?
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