实例讲解Python中函数的调用与定义

程序员文章站 2022-05-06 13:31:06

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调用函数：

#!/usr/bin/env python3 
# -*- coding: utf-8 -*- 
 
# 函数调用 
>>> abs(100) 
100 
>>> abs(-110) 
110 
>>> abs(12.34) 
12.34 
>>> abs(1, 2) 
Traceback (most recent call last): 
 File "", line 1, in  
TypeError: abs() takes exactly one argument (2 given) 
>>> abs('a') 
Traceback (most recent call last): 
 File "", line 1, in  
TypeError: bad operand type for abs(): 'str' 
>>> max(1, 2) 
2 
>>> max(2, 3, 1, -5) 
3 
>>> int('123') 
123 
>>> int(12.34) 
12 
>>> str(1.23) 
'1.23' 
>>> str(100) 
'100' 
>>> bool(1) 
True 
>>> bool('') 
False 
>>> a = abs # 变量a指向abs函数，相当于引用 
>>> a(-1) # 所以也可以通过a调用abs函数 
1 
 
>>> n1 = 255 
>>> n2 = 1000 
>>> print(hex(n1)) 
0xff 
>>> print(hex(n2)) 
0x3e8

定义函数：

#!/usr/bin/env python3 
# -*- coding: utf-8 -*- 
 
#函数定义 
def myAbs(x): 
 if x >= 0: 
  return x 
 else: 
  return -x 
 
a = 10 
myAbs(a) 
 
def nop(): # 空函数 
 pass

pass语句什么都不做。
实际上pass可以用来作为占位符，比如现在还没想好怎么写函数代码，就可以先写一个pass，让代码运行起来。

if age >= 18: 
 pass 
#缺少了pass，代码就会有语法错误 
>>> if age >= 18: 
... 
 File "", line 2 
 
 ^ 
IndentationError: expected an indented block 
 
>>> myAbs(1, 2) 
Traceback (most recent call last): 
 File "", line 1, in  
TypeError: myAbs() takes 1 positional argument but 2 were given 
>>> myAbs('A') 
Traceback (most recent call last): 
 File "", line 1, in  
 File "", line 2, in myAbs 
TypeError: unorderable types: str() >= int() 
>>> abs('A') 
Traceback (most recent call last): 
 File "", line 1, in  
TypeError: bad operand type for abs(): 'str' 
 
def myAbs(x): 
 if not isinstance(x, (int, float)): 
  raise TypeError('bad operand type') 
 if x >= 0: 
  return x 
 else: 
  return -x 
 
>>> myAbs('A') 
Traceback (most recent call last): 
 File "", line 1, in  
 File "", line 3, in myAbs 
TypeError: bad operand type

返回两个值？

import math 
def move(x, y, step, angle = 0): 
 nx = x + step * math.cos(angle) 
 ny = y - step * math.sin(angle) 
 return nx, ny 
 
>>> x, y = move(100, 100, 60, math.pi / 6) 
>>> print(x, y) 
151.96152422706632 70.0

其实上面只是一种假象，Python函数返回的仍然是单一值。

>>> r = move(100, 100, 60, math.pi / 6) 
>>> print(r) 
(151.96152422706632, 70.0)

实际上返回的是一个tuple！
但是，语法上，返回一个tuple可以省略括号，而多个变量可以同时接受一个tuple，按位置赋给对应的值。
所以，Python的函数返回多值实际就是返回一个tuple，但是写起来更方便。
函数执行完毕也没有return语句时，自动return None。

练习：

import math 
def quadratic(a, b, c): 
 x1 = (-b + math.sqrt(b * b - 4 * a * c)) / (2 * a) 
 x2 = (-b - math.sqrt(b * b - 4 * a * c)) / (2 * a) 
 return x1, x2 
 
x1, x2 = quadratic(2, 5, 1) 
print(x1, x2) 
 
>>> import math 
>>> def quadratic(a, b, c): 
...  x1 = (-b + math.sqrt(b * b - 4 * a * c)) / (2 * a) 
...  x2 = (-b - math.sqrt(b * b - 4 * a * c)) / (2 * a) 
...  return x1, x2 
... 
>>> x1, x2 = quadratic(2, 5, 1) 
>>> print(x1, x2) 
-0.21922359359558485 -2.2807764064044154