菜单类，压栈，出栈，查询功能

步骤：

1.程序运行是不是交互的，主功能的需要的位置参数

2.菜单类，压栈，出栈，查询功能

3.先写框架，先后执行的顺序，写函数便于简化，主程序的运行

4.push pop view main框架完善，调入模块，定义变量

def push_it():

def pop_it():

def view_it():

def show_menu():
    
if __name__ == '__main__':
    show_menu()

改进

def push_it():
    print('push')

def pop_it():
    print('pop')

def view_it():
    print('view')

def show_menu():
    prompt = """(0)push it
(1)pop it
(2)view it
(3)exit
Please input you choice(0/1/2/3):"""

    while True:
        choice = input(prompt).strip()[0]
        if choice not in '0123':
            print('Invalid input. Try again.')
            continue

        if choice == '3':
            break
        if choice == '0':
            push_it()
        elif choice == '1':
            pop_it()
        else:
            view_it()

if __name__ == '__main__':
    show_menu()

#############################################################

stack = []


def push_it():
    item = input('item to push :')
    stack.append(item)


def pop_it():
    if stack:
        print('from stack popped %s'  % stack.pop())


def view_it():
    print(stack)


def show_menu():
    cmds = {'0': push_it, '1': pop_it, '2': view_it}
    prompt = """(0)push it
(1)pop it
(2)view it
(3)exit
Please input you choice(0/1/2/3):"""

    while True:
        choice = input(prompt).strip()[0]

        if choice not in '0123':
            print('Invalid input. Try again.')
            continue

        if choice == '3':
            break

        cmds[choice]()


if __name__ == '__main__':
    show_menu()