摘要: def count(): fs = [] for i in range(1, 4): def f(): return i*i fs.append(f) return fsf1,f2,f3 = count()print(f1()) #等到i = 3 了,才return fs统一返回f函数的3个对象赋值 阅读全文
posted @ 2018-07-25 17:14 只记今朝笑 阅读(552) 评论(0) 推荐(0) 编辑