PythonStudy——函数对象的案例

# part1

# 加法运算

def add(n1, n2):
    return n1 + n2

def low(n1, n2):
    return n1 - n2

# 四则运算

def computed(n1, n2, func):
    # if cmd == 'add':
    #     return add(n1, n2)
    # elif cmd == 'low':
    #     return n1 - n2
    return func(n1, n2)

r1 = computed(10, 20, add)
print(r1)
r2 = computed(10, 20, low)
print(r2)

# part2

def add(n1, n2):
    return n1 + n2

def low(n1, n2):
    return n1 - n2

def computed(n1, n2, func):
    return func(n1, n2)

# 测试
cmd = input('cmd: ')  # 只能等于字符串add、low => 什么方式可以将字符串add、low对应上函数
fn = None
if cmd == 'add':
    fn = add
elif cmd == 'low':
    fn = low
computed(10, 20, fn)
# 直接输入cmd 也及时传入了一个函数对象

# 测试
fn_map = {
    'add': add,
    'low': low
}

if cmd in fn_map:  # 作为容器的成员可以简化判断逻辑
    fn = fn_map[cmd]
    res = computed(10, 20, fn)
    print(res)
else:
    print('该运算暂不支持')

# part 3

def add(n1, n2):
    return n1 + n2

def low(n1, n2):
    return n1 - n2

def jump(n1, n2):
    return n1 * n2

def computed(n1, n2, func):
    return func(n1, n2)
fn_map = {
    'add': add,
    'low': low,
    'jump': jump
}

def get_fn(cmd):
    f = add  # 默认为add函数
    if cmd in fn_map:
        f = fn_map[cmd]  # 如果指令正确就返回对应的函数，如果不正确，就是采用默认的函数
    return f


while True:
     cmd = input('cmd: ')
     if cmd == 'q':
         break
     if cmd in fn_map:
         fn = fn_map[cmd]  # 通过指令找函数
         res = computed(10, 20, fn)
         print(res)
     else:
         print('该运算暂不支持')