@author jackzhenguo @desc @date 2019/2/20
21 nonlocal用于内嵌函数中
关键词nonlocal常用于函数嵌套中，声明变量i为非局部变量；
如果不声明，i+=1表明i为函数wrapper内的局部变量，因为在i+=1引用(reference)时,i未被声明，所以会报unreferenced variable的错误。
def excepter(f): i = 0 t1 = time.time() def wrapper(): try: f() except Exception as e: nonlocal i i += 1 print(f'{e.args[0]}: {i}') t2 = time.time() if i == n: print(f'spending time:{round(t2-t1,2)}') return wrapper