Generator usage in Python

Question

I'm learning about generators, so I defined the following function which calculates Fibonacci sequence:

def fib(max):
    a, b = 0, 1
    while a < max:
        yield a
        a, b = b, a + b

I tried to use it like this, but it didn't work:

next(fib(10))
Out[413]: 0

next(fib(10))
Out[414]: 0

next(fib(10))
Out[415]: 0

However, using it like this works as expected:

f = fib(10)

next(f)
Out[417]: 0

next(f)
Out[418]: 1

next(f)
Out[419]: 1

next(f)
Out[420]: 2

Why the first case does not work?

Michal Frystacky · Accepted Answer · 2016-02-15 04:58:49Z

2

next(iterator[, default])

Retrieve the next item from the iterator by calling its next() method. If default is given, it is returned if the iterator is exhausted, otherwise StopIteration is raised.

The first one works:

It just generates a new iterator for each of the functions instances you make when you callfib(10)

So, every time you call fib(10) you create a new fib function instance that returns an iterator specific to that instance.

Notice that they all return the first value correctly.

edited Feb 15, 2016 at 4:58

answered Feb 15, 2016 at 4:52

Michal Frystacky

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4 Comments

MichaelSB Over a year ago

Oh, so calling fib(10) several times is not the same as calling f several times?

Michal Frystacky Over a year ago

Every time you call fib(10) you are calling an entirely new function, but when you call f you are calling a single instance of fib(10)

MichaelSB Over a year ago

Ok, that explains it.

Michal Frystacky Over a year ago

@MichaelSB I am glad it made sense!

Seekheart · Accepted Answer · 2016-02-15 04:53:18Z

0

when you run it in the first case it starts again from the beginning and so it'll always just go start --> first, while in the second case everytime you call f, f is changing so it goes start --> first --> second --> third.

answered Feb 15, 2016 at 4:53

Seekheart

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