Small turtle Lesson 19: Functions: I listen to my site after-school summary

0. What would enter the following program?

def next():
    print('我在next()函数里...')
    pre()
def pre():
    print('我在pre()函数里...')
    
next()

我在next()函数里...
我在pre()函数里...

Summary: In which joined the pre def next, but also pre defined below, it will all be printed.

4. The following visual program will print what?

var = ' Hi '
def fun1():
    global var
    var = ' Baby '
    return fun2(var)
def fun2(var):
    var += 'I love you'
    fun3(var)
    return var
def fun3(var):
    var = ' 小甲鱼 '
print(fun1())

Var 1.global the first to define a global variable, so hi is changed, if at this time the last line plus print (var). It will print baby. If you remove the global, after a line is printed together with Hi
2.fun3 and not return a value, so that coding such as the following:

var = 'hi'
#注意这里一定要定义！因为在函数中的为局部变量
def fun3(var):
    var = ' 小甲鱼 '
fun3(var)

It will print more empty, because there is no value to the fun3, and so on, so the source code fun3 invalid
value so the final will be printed: Baby i love you

0. write a function that determines whether the passed string argument "palindrome associated" (i.e., associated with a palindromic palindromic written in the form of couplets, can read along, for example, may be read backwards: Shanghai water from the sea).

自己的答案（错误！）
def back():
    print('请输入查询：',end='')
    temp = input()
    i = 0
    z = 0
    while i < (len(temp)//2):
        temp1 = temp[i]
        if temp.find(temp1,i,i) == temp.rfind(temp1,i,i):
            z = 2
            i +=1
        else:
            z = 1
            break
    if z == 1 :
        print('不是')
    elif z == 2 :
        print('是')

back()

find, rfind usage misunderstanding: the following example

str_super = "this is really a string example....wow!!!"
substr = "s"

print(str_super.rfind(substr))
print(str_super.rfind(substr, 0, 10))
print(str_super.rfind(substr, 10, 0))

print(str_super.find(substr))
print(str_super.find(substr, 0, 10))
print(str_super.find(substr, 10, 0))
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版权声明：本文为CSDN博主「追风筝的人000」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/weixin_44901453/article/details/89598154
结果：
17
6
-1
3
3
-1

You can see, rfind role is to look backwards to the first s, then the output location, location and location for the same number of forward
if there is a large interest in God can help me rephrase the above procedure makes it feasible.

网上答案一：https://www.jianshu.com/p/f9bb090aaf24
def palindrome(string):
    length = len(string)
    last = length-1
    length //= 2
    flag = 1
    for each in range(length)
    #注意:range一个数是从0到这个数减一，不包括它本身
        if string[each] != string[last]:
            flag = 0
        last -= 1
    if flag == 1:
        return 1
    else:
        return 0
string = input('请输入一句话：')
if palindrome(string) == 1:
    print('是回文联!')
else:
    print('不是回文联！')

The second was the use of a reverse list, it is necessary to advance the string becomes list form

def back():
    list1 = []
    list2 = []
    print('请输入字符串：',end='')
    temp = input()
    list1 = list(temp)
    list2 = reversed(list1)

    if list1 == list(list2):
        print('是回文联')
    else:
        print('八是哦')

back()

Here we must pay attention to the issue of reverse, and in writing this code, when I first used the list.reverse list (), and then assigned to list2.
The results showed that output list2 is empty, while list1 been reversed.

def back():
    list1 = []
    list2 = []
    print('请输入字符串：',end='')
    temp = input()
    list1 = list(temp)
    list2 = list1.reverse()

    if list1 == list2:
        print('是回文联')
    else:
        print('八是哦')
back()

结果：
请输入字符串：asddddd
['d', 'd', 'd', 'd', 'd', 's', 'a']
None
八是哦

This is because the reverse is list1 built-in functions, in order to save time, so directly list1 changes, due to the return value is null.
Therefore, if you want to assign to list here, but only to take a sequence operator reversed, and the list again in order to get the desired value.

1. Write a function, respectively, the statistics of the number of incoming string parameter (may be more than one parameter) of letters, spaces, numbers and other characters.

def number(*s):
    length = len(s)
    for z in range(length):
       i = 0
       a = 0
       b = 0
       c = 0
       for each in s[z]:
           if each.isdecimal():
               i +=1
           elif each.isalpha():
               a +=1
           elif each.isspace():
               b +=1
           else:
               c +=1
       print('第',z,'段字符含有数字：',i,'个 英文：',a,'个 空格：',b,'个 特殊字符：',c,'个')


number ('I love fish.com 123', 'I love you', 'you love 123')
    

note! ! ! ! ! 1. String built-in functions remember playing parentheses! ! ! !
2. If the input with the input 'I love fish.com 123', ' I love you', 'you love 123', the default input to a string, and not as above in which three strings numer ** (here to be understood!) **