实例讲解Python中函数的调用与定义
发布时间:2020-12-16 20:52:53 所属栏目:Python 来源:网络整理
导读: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
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>",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 。 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,in <module> TypeError: myAbs() takes 1 positional argument but 2 were given >>> myAbs('A') Traceback (most recent call last): File "<stdin>",in <module> File "<stdin>",line 2,in myAbs TypeError: unorderable types: str() >= int() >>> abs('A') Traceback (most recent call last): File "<stdin>",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 3,in myAbs TypeError: bad operand type 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 >>> r = move(100,math.pi / 6) >>> print(r) (151.96152422706632,70.0) 实际上返回的是一个tuple! 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,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 ... >>> x1,1) >>> print(x1,x2) -0.21922359359558485 -2.2807764064044154 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |