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