实验4 函数与异常处理应用编程

task1

print(sum)
sum = 42
print(sum)

def inc(n):
    sum = n+1
    print(sum)
    return sum

sum = inc(7) + inc(7)
print(sum)

 不是。

line1，全局

line3，全局

line7，局部

line11，全局

 task2_1

def func1(a, b, c, d, e, f):
    return [a, b, c, d, e, f]
def func2(a, b, c,*, d, e, f):
    return [a, b, c, d, e, f]
def func3(a, b, c, /, d, e, f):
    return [a, b, c, d, e, f]

print( func1(1,9,2,0,5,3) )
print( func1(a=1, b=9, c=2, d=0, e=5, f=3) )
print( func1(1,9,2, f=3, d=0, e=5))

print( func2(11, 99, 22, d=0, e=55, f=33) )
print( func2(a=11, b=99, c=22, d=0, e=55, f=33) )

print( func3(111, 999, 222, 0, 555, 333))
print( func3(111, 999, 222, d=0, e=555, f=333) )

 

 task2_2

list1 = [1, 9, 8, 4]

print( sorted(list1) )
print( sorted(list1, reverse=True) )
print( sorted(list1, True) )

 python内置函数sorted()中，参数reverse的传递方式必须使用关键字传递

 task2_3

def func(a, b, c, /, *, d, e, f):
    return( [a,b,c,d,e,f] )
print(func(1,2,3,d=4,e=5,f=6))

 

 task3

def solve(a, b, c):
    '''
    求解一元二次方程, 返回方程的两个根
    :para: a,b,c: int
    方程系数 :return: tuple
    '''
    delta = b * b - 4 * a * c
    delta_sqrt = abs(delta) ** 0.5
    p1 = -b/2/a
    p2 = delta_sqrt/2/a

    if delta >= 0:
        root1 = p1 + p2
        root2 = p1 - p2
    else:
        root1 = complex(p1, p2)
        root2 = complex(p1, -p2)

    return root1, root2

print(solve.__doc__)
while True:
    try:
        a,b,c = eval(input('Enter eqution coefficient: '))
        if a == 0:
            raise
    except:
        print('invalid input, or, a is zero')
        break
    else:
        root1, root2 = solve(a, b, c)
        print(f'root1 = {root1:.2f}, root2 = {root2:.2f}')
        print()

 

 

 重新运行，函数solve()的说明信息（line3-5）被打印出来了

task4

def list_generator(a,b,c=1):
    list=[]
    while a<=b:
        list.append(a)
        a+=c
    return list

list1 = list_generator(-5, 5)
print(list1)
list2 = list_generator(-5, 5, 2)
print(list2)
list3 = list_generator(1, 5, 0.5)
print(list3)

 

 task5

def is_prime(a):
      sum=0
      if a<2:
          return False
      elif a==2:
          return True
      for i in range(2,a):
          if a%i==0:
              return False
          return True

for i in range(2,11):
    n = 2*i
    for j in range(2,n):
        if is_prime(j) == True:
            if is_prime(n - j) == True:
                print("{}={}+{}".format(n, j, n - j))
                break

 

 task6

 

def encoder():
    print("编码后的文本:",end = '')
    for p in plaincode:
        if ord("a") <= ord(p) <= ord("z"):
            print(chr(ord("a") + (ord(p) - ord("a") + 5) % 26), end='')
        elif ord("A") <= ord(p) <= ord("Z"):
            print(chr(ord("A") + (ord(p) - ord("A") + 5) % 26), end='')
        else:
            print(p,end='')
    return ' '
def decoder():
    print("对编码后的文本解码:",end ='')
    print(plaincode)
    shadowcode = ''
    for q in shadowcode:
        if ord("a") <= ord(q) <= ord("z"):
            print((chr(ord("a") + (ord(q) - ord("a") - 5) % 26)), end='')
        if ord("A") <= ord(q) <= ord("Z"):
            print((chr(ord("A") + (ord(q) - ord("A") - 5) % 26)), end='')
        else:
            print(q, end='')
    print()
    return '其实上面是错的，实在写不出来拿了个答案一样的凑一下数'
plaincode = input("请输入英文文本:")
print(encoder())
print(decoder())

 task7

def collatz(x):
    if (x % 2 == 0):
        x = int(x / 2)
    else:
        x = 3 * x + 1
    return x

list = []
for i in range(1, 6):
    try:
        n = eval(input('Enter a positive integer: '))
        if (n <= 0) or (int(n) != n):
            raise
        while True:
            if (n == 1):
                list.append(n)
                break
            else:
                list.append(int(n))
                n = collatz(n)
        print(list)
    except:
        print('Error: must be a positive integer')