__next__和__iter__实现迭代器协议

__next__和__iter__实现迭代器协议

#_*_coding:utf-8_*_
__author__ = 'Linhaifeng'
class Foo:
    def __init__(self,x):
        self.x=x

    def __iter__(self):
        return self

    def __next__(self):
        n=self.x
        self.x+=1
        return self.x

f=Foo(3)
for i in f:
    print(i)
简单示范
class Foo:
    def __init__(self,start,stop):
        self.num=start
        self.stop=stop
    def __iter__(self):
        return self
    def __next__(self):
        if self.num >= self.stop:
            raise StopIteration
        n=self.num
        self.num+=1
        return n

f=Foo(1,5)
from collections import Iterable,Iterator
print(isinstance(f,Iterator))

for i in Foo(1,5):
    print(i) 
class Range:
    def __init__(self,n,stop,step):
        self.n=n
        self.stop=stop
        self.step=step

    def __next__(self):
        if self.n >= self.stop:
            raise StopIteration
        x=self.n
        self.n+=self.step
        return x

    def __iter__(self):
        return self

for i in Range(1,7,3): #
    print(i)
练习:简单模拟range,加上步长
class Fib:
    def __init__(self):
        self._a=0
        self._b=1

    def __iter__(self):
        return self

    def __next__(self):
        self._a,self._b=self._b,self._a + self._b
        return self._a

f1=Fib()

print(f1.__next__())
print(next(f1))
print(next(f1))

for i in f1:
    if i > 100:
        break
    print('%s ' %i,end='')
斐波那契数列
posted @ 2019-05-07 23:04  樵夫-justin  阅读(102)  评论(0)    收藏  举报