实例讲解Python中函数的调用与定义


            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 "<stdin>", line 1, in <module> 

TypeError: abs() takes exactly one argument (2 given) 

>>> abs('a') 

Traceback (most recent call last): 

 File "<stdin>", line 1, in <module> 

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 "<stdin>", line 2 

 

 ^ 

IndentationError: expected an indented block 

 

>>> myAbs(1, 2) 

Traceback (most recent call last): 

 File "<stdin>", line 1, in <module> 

TypeError: myAbs() takes 1 positional argument but 2 were given 

>>> myAbs('A') 

Traceback (most recent call last): 

 File "<stdin>", line 1, in <module> 

 File "<stdin>", line 2, in myAbs 

TypeError: unorderable types: str() >= int() 

>>> abs('A') 

Traceback (most recent call last): 

 File "<stdin>", line 1, in <module> 

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 "<stdin>", line 1, in <module> 

 File "<stdin>", 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