Next() function.

Question

def numbers(x):
for i in range(x):
if i%2==0:
yield i
value=numbers(11)
print(next(value))
print(next(value))
print(next(value))
print(next(value))
print(next(value))

>this program gives output as 0,2,4,6,8,10
but ,

def numbers(x):
for i in range(x):
if i%2==0:
yield i
print(next(numbers(11))
print(next(numbers(11))
print(next(numbers(11))
print(next(numbers(11))
print(next(numbers(11))
print(next(numbers(11))
>this program always give output 0 ..why??

Answer 1

In 2nd program , you are calling the next function only once on each function unlike the first one that has multiple next call on same function .

Answer 2

your function define/return a generator (because use of 'yield' keyword inside)...

then you must store it in a variable and call next(generator) to get each values returned by 'yield' statement(s) ^^

in the second case, you build a new generator at each next call, so you only get the first value returned by 'yield' at each call ;P