Unreachable 'yield' still strangely affects function [duplicate]

Question

If you run the following snippet, it will ignore the 'print' on line 2 and 'exit' on line 3. However, if you comment out the already unreachable 'yield' from line 4, lines 2 and 3 will execute normally.

This makes me think that Python (3.5.2) looks for a 'yield' anywhere in the function, and if one is found, even an unreachable one, nothing is executed until a next() is called on the returned iterator. Up until now, I was under the impression that the function would execute normally up until hitting a yield, at which point it would begin acting like an iterator.

def func():
    print("> Why doesn't this line print?")
    exit() # Within this function, nothing should matter after this point.  The program should exit
    yield "> The exit line above will exit ONLY if you comment out this line."


x = func()
print(x)

Does this seem strange to anyone else? Does anybody have some insight to share here?

Answer 1

When a function has a yield, even if it is not reachable, it will be a generator.

Answer 2

The thing here is the yield statement. When you use yield inside a function, it will create a generator. Keep in mind that any function with a yield will return a generator when called.

Try this:

def func():
    print("> Why doesn't this line print?")
    exit()

x = func()
print(x)

Results:

Why doesn't this line print?
None

And then the program exits.