python的函数能直接调用外部的变量，所以有些时候没必要传参了？

[baidu_28726667]

函数内是可以直接打印外面的变量？！ 但是却不能运算？！

import os

def printdrink():
    print("drink:", drink[i])
    #test+=1 会报错，屏蔽
    print(test)


test = 111

drink =['coffee','black_tea','milk_tea','cococa','water']
print("type drink: ",type(drink))
for i in range(len(drink)):
    printdrink()

结果

type drink: <class 'list'>
drink: coffee
111
drink: black_tea
111
drink: milk_tea
111
drink: cococa
111
drink: water
111

另外这个test在函数内并不能 test+=1 会报错 local variable 'test' referenced before assignment ？！ 意思说我使用了一个没初始值的局部变量 ？

如果代码改为

import os

def printdrink(test):
    print("drink:", drink[i])
    test+=1
    print(test)


test = 111

drink =['coffee','black_tea','milk_tea','cococa','water']
print("type drink: ",type(drink))
for i in range(len(drink)):
    printdrink(test)

这样函数里面永远都是打印112.这代表了这个test只是值传递。有没有办法强制引用传递呢？

[ruancan]

你应该是只想让test变量随着printdrink调用次数而增加。

def printdrink():
    print("drink:", drink[i])
    global test
    test += 1
    print(test)


test = 111

drink = ['coffee', 'black_tea', 'milk_tea', 'cococa', 'water']
print("type drink: ", type(drink))
for i in range(len(drink)):
    printdrink()

[YXTS122]