我试图在Python中实现方法重载:

class A:
    def stackoverflow(self):    
        print 'first method'
    def stackoverflow(self, i):
        print 'second method', i

ob=A()
ob.stackoverflow(2)

但是输出是第二种方法2;类似的:

class A:
    def stackoverflow(self):    
        print 'first method'
    def stackoverflow(self, i):
        print 'second method', i

ob=A()
ob.stackoverflow()

给了

Traceback (most recent call last):
  File "my.py", line 9, in <module>
    ob.stackoverflow()
TypeError: stackoverflow() takes exactly 2 arguments (1 given)

我该怎么做呢?


当前回答

我刚刚遇到了重载.py (Python 3的函数重载),有兴趣的同学可以去看看。

从链接库的README文件:

overloading is a module that provides function dispatching based on the types and number of runtime arguments. When an overloaded function is invoked, the dispatcher compares the supplied arguments to available function signatures and calls the implementation that provides the most accurate match. Features Function validation upon registration and detailed resolution rules guarantee a unique, well-defined outcome at runtime. Implements function resolution caching for great performance. Supports optional parameters (default values) in function signatures. Evaluates both positional and keyword arguments when resolving the best match. Supports fallback functions and execution of shared code. Supports argument polymorphism. Supports classes and inheritance, including classmethods and staticmethods.

其他回答

我用Python 2.7写了我的答案:

在Python中,方法重载是不可能的;如果你真的想用不同的特性访问同一个函数,我建议你使用方法重写。

class Base(): # Base class
    '''def add(self,a,b):
        s=a+b
        print s'''

    def add(self,a,b,c):
        self.a=a
        self.b=b
        self.c=c

        sum =a+b+c
        print sum

class Derived(Base): # Derived class
    def add(self,a,b): # overriding method
        sum=a+b
        print sum



add_fun_1=Base() #instance creation for Base class
add_fun_2=Derived()#instance creation for Derived class

add_fun_1.add(4,2,5) # function with 3 arguments
add_fun_2.add(4,2)   # function with 2 arguments

我想你想说的是"超载"Python中没有任何方法重载。但是,您可以使用默认参数,如下所示。

def stackoverflow(self, i=None):
    if i != None:
        print 'second method', i
    else:
        print 'first method'

当您向它传递一个参数时,它将遵循第一个条件的逻辑并执行第一个print语句。当你不给它传递参数时,它将进入else条件并执行第二个print语句。

在MathMethod.py文件中:

from multipledispatch import dispatch
@dispatch(int, int)
def Add(a, b):
   return a + b 
@dispatch(int, int, int)  
def Add(a, b, c):
   return a + b + c 
@dispatch(int, int, int, int)    
def Add(a, b, c, d):
   return a + b + c + d

在Main.py文件

import MathMethod as MM 
print(MM.Add(200, 1000, 1000, 200))

我们可以使用multipledispatch重载该方法。

我刚刚遇到了重载.py (Python 3的函数重载),有兴趣的同学可以去看看。

从链接库的README文件:

overloading is a module that provides function dispatching based on the types and number of runtime arguments. When an overloaded function is invoked, the dispatcher compares the supplied arguments to available function signatures and calls the implementation that provides the most accurate match. Features Function validation upon registration and detailed resolution rules guarantee a unique, well-defined outcome at runtime. Implements function resolution caching for great performance. Supports optional parameters (default values) in function signatures. Evaluates both positional and keyword arguments when resolving the best match. Supports fallback functions and execution of shared code. Supports argument polymorphism. Supports classes and inheritance, including classmethods and staticmethods.

在3.4之前,agf的答案是正确的,而现在有了PEP-3124,我们得到了语法糖。

有关@overload装饰器的详细信息,请参阅typing文档,但请注意,这真的只是语法糖,恕我冒犯,这是人们一直在争论的问题。

就我个人而言,我同意拥有多个具有不同签名的函数比拥有一个具有20多个参数都设置为默认值(大多数时候没有)的单一函数更具可读性,然后不得不使用无休止的if, elif, else链来找出调用者实际希望我们的函数对所提供的参数集做什么。在Python禅之后,这是早就该有的:

美总比丑好。

也可以说

简单比复杂好。

直接来自上面链接的Python官方文档:

from typing import overload
@overload
def process(response: None) -> None:
    ...
@overload
def process(response: int) -> Tuple[int, str]:
    ...
@overload
def process(response: bytes) -> str:
    ...
def process(response):
    <actual implementation>

编辑:如果有人想知道为什么这个例子不能像你从其他语言中期望的那样工作,我建议看看这个讨论。@overloaded函数不应该有任何实际的实现。这在Python文档中的示例中并不明显。