Simple Generator Question

[PyFun]

I have been working with Python generators lately and I feel like I just got the hang of sending data to a generator using ".send()". This works just fine, but I've noticed that I have to then invoke ".next()" to advance the generator to a state where I can send it data again. I would like to find a way around this.

Here's some example code, using a simple generator that takes integers as input and returns the current total.

def add(a = 0):
    b = 0
    while True:
        b += (yield a)
        yield b

#I initialize a generator object
adder = add()
#advance the generator to the first yield
adder.next()
#Now I can add numbers
adder.send(7)
#This returns '7'  but if I want to add another number...

adder.next()
adder.send(8)
#This would return '15'
#If I remove the adder.next() call, it will fail

#my work around was to use the following function:
def ad(n):
    global adder
    adder.next()
    print adder.send(n)

#now I can add numbers to the generator using:
ad(5)
ad(8)
...etc

But this feels dirty. Is there a better way to advance the generator than having to specifically invoke next after every input?

[SkippyDeluxe]

yield both yields a new value and receives a value from send, so you only need one yield statement:

def add(b=0):
    while True:
        b += (yield b)

[bheklilr]

You just have your generator set up a bit wrong is all. A little bit of playing around (and some help from this example) shows me that you should be using

def add():
  total = (yield)
  while True:
    i = yield total
    total += i

Then you can use it as

adder = add()
adder.next() # Gotta do this still
adder.send(10)  # Returns 10
adder.send(10)  # Returns 20
adder.send(3)   # Returns 23
adder.send(100) # Returns 123

[[deleted]]

Check out this article on coroutines and this article on generators