【参考廖雪峰Python教程——https://www.liaoxuefeng.com/wiki/1016959663602400/1017434209254976】
【自学使用,请勿转载】
返回函数返回值是函数
def lazy_sum(*args): def sum(): ax = 0 for n in args: ax = ax + n return ax return sumf = lazy_sum(1, 3, 5, 7, 9)print(f())
【结果】:25
f是函数名
f()是调用函数
【调试检查】:
def lazy_sum(*args): def sum(): ax = 0 for n in args: ax = ax + n return ax print(sum) return sumf = lazy_sum(1, 3, 5, 7, 9)print(f)
【结果】:
<function lazy_sum.<locals>.sum at 0x000001DB164D2700>
<function lazy_sum.<locals>.sum at 0x000001DB164D2700>
且调用函数时,即使传入相同的参数,返回的函数也并不相同。
def lazy_sum(*args): def sum(): ax = 0 for n in args: ax = ax + n return ax return sum f1 = lazy_sum(1, 3, 5, 7, 9)f2 = lazy_sum(1, 3, 5, 7, 9)print(f1==f2)
【结果】:False
闭包对于lazy_sum(*args)函数来说,*args参数是局部变量,但返回的sum由于在定义内部使用了args,所以相关的参数和变量(args的相关参数和变量)都保存在返回的函数(sum)中。
def lazy_sum(*args): def sum(): ax = 0 for n in args: ax = ax + n return ax return sum
【注意】:返回的函数没有立刻执行,而是等到调用的时候才执行!
def count(): fs = [] for i in range(1, 4): def f(): return i*i fs.append(f) return fsf1, f2, f3 = count()print(f1(), f2(), f3())
【结果】:9 9 9
结果并非是1 4 9而是9 9 9,为什么呢?
【调试检查】:
def count(): fs = [] for i in range(1, 4): def f(): return i*i fs.append(f) print(fs) return fsf1, f2, f3 = count()print(f1,',',f2,',', f3)print(f1(), f2(), f3())
【结果】:
[<function count.<locals>.f at 0x00000153A4F73700>, <function count.<locals>.f at 0x00000153A4F73790>, <function count.<locals>.f at 0x00000153A4F73820>]
<function count.<locals>.f at 0x00000153A4F73700> , <function count.<locals>.f at 0x00000153A4F73790> , <function count.<locals>.f at 0x00000153A4F73820>
9 9 9
可以看出每次循环都会创建新的函数加入fs列表中,但它们并不是立即执行,它们其实是在调用的时候才执行,而在调用的时候,返回函数引用的变量i已经等于3了,所以,结果就是9 9 9
因此,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!
如果一定要引用循环变量,那就再加一个函数。
def count(): fs = [] for i in range(1, 4): def f(j): def g(): return j*j return g fs.append(f(i)) return fsf1, f2, f3 = count()print(f1(), f2(), f3())
【结果】:1 4 9
【调试检查】:
def count(): fs = [] for i in range(1, 4): def f(j): def g(): return j*j return g fs.append(f(i)) print(fs) return fsf1, f2, f3 = count()print(f1,',',f2,',', f3)print(f1(), f2(), f3())
【结果】:
[<function count.<locals>.f.<locals>.g at 0x00000237F4353790>, <function count.<locals>.f.<locals>.g at 0x00000237F4353820>, <function count.<locals>.f.<locals>.g at 0x00000237F43538B0>]
<function count.<locals>.f.<locals>.g at 0x00000237F4353790> , <function count.<locals>.f.<locals>.g at 0x00000237F4353820> , <function count.<locals>.f.<locals>.g at 0x00000237F43538B0>
1 4 9
由于f(i)被立刻执行,返回函数引用的变量j在加进列表的时候就封装进函数了。
总结以上是内存溢出为你收集整理的Python学习之【返回函数】全部内容,希望文章能够帮你解决Python学习之【返回函数】所遇到的程序开发问题。
如果觉得内存溢出网站内容还不错,欢迎将内存溢出网站推荐给程序员好友。
欢迎分享,转载请注明来源:内存溢出
评论列表(0条)