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Copy pathreturn_func.py
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113 lines (78 loc) · 1.93 KB
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#!/bin/env python3
# -*- coding:utf-8 -*-
# 返回求和函数
def lazy_sum(*args):
def sum3():
ax = 0
for n in args:
ax = ax + n
return ax
return sum3
# 调用返回函数,注意添加函数括号
print(lazy_sum(1, 3, 5, 7, 9))
print(lazy_sum(1, 3, 5, 7, 9)())
# 调用返回函数，相同参数返回函数都不同，互不影响
f1 = lazy_sum(1, 3, 5, 7, 9)
f2 = lazy_sum(1, 3, 5, 7, 9)
print(f1 == f2)
# why f1(), f2(), f3() returns 9, 9, 9 rather than 1, 4, 9?
# 返回函数并非立即执行，返回函数不要引用任何循环变量，或者后续会发生变化的变量。
def count1():
fs = []
for i in range(1, 4):
def f():
return i * i
fs.append(f)
return fs # 迭代完f()执行完后返回fs,fs为9
f1, f2, f3 = count1()
print(f1(), f2(), f3())
# 返回函数中引用循环变量
def count2():
def f(j):
def g():
return j * j
return g
fs = []
for i in range(1, 4):
fs.append(f(i)) # f(i)立刻被执行，因此i的当前值被传入f()
return fs
f1, f2, f3 = count2()
print(f1(), f2(), f3())
# use generator
def createCounter1():
def f():
n = 0
while True:
n += 1
yield n # 先创造一个生成器
sum1 = f()
def counter():
return next(sum1) # 用一个函数来调用生成器
return counter
M1 = createCounter1()
print(M1()) # 注意这里的调用返回函数
print(M1())
print(M1())
# use list
def createCounter2():
fs = [0]
def counter():
fs[0] = fs[0] + 1
return fs[0]
return counter
M2 = createCounter2()
print(M2())
print(M2())
print(M2())
# 使用nonlocal关键字，将局部变量变成全局变量
def createCounter3():
i = 0
def counter():
nonlocal i
i += 1
return i
return counter
M3 = createCounter3()
print(M3())
print(M3())
print(M3())