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

其他回答

你也可以使用pythonlangutil:

from pythonlangutil.overload import Overload, signature

class A:
    @Overload
    @signature()
    def stackoverflow(self):    
        print('first method')
    
    @stackoverflow.overload
    @signature("int")
    def stackoverflow(self, i):
        print('second method', i)

Python不像Java或c++那样支持方法重载。我们可以重载方法,但只能使用最新定义的方法。

# First sum method.
# Takes two argument and print their sum
def sum(a, b):
    s = a + b
    print(s)

# Second sum method
# Takes three argument and print their sum
def sum(a, b, c):
    s = a + b + c
    print(s)

# Uncommenting the below line shows an error
# sum(4, 5)

# This line will call the second sum method
sum(4, 5, 5)

我们需要提供可选参数或*args,以便在调用时提供不同数量的参数。

方法重载

在Python中,重载不是一个应用概念。然而,如果你试图创建这样的情况,例如,你想要在传入类型为foo的参数时执行一个初始化式,而在传入类型为bar的参数时执行另一个初始化式,那么,由于Python中的所有内容都是作为对象处理的,你可以检查传入对象的类类型的名称,并基于此编写条件处理。

class A:
   def __init__(self, arg)
      # Get the Argument's class type as a String
      argClass = arg.__class__.__name__

      if argClass == 'foo':
         print 'Arg is of type "foo"'
         ...
      elif argClass == 'bar':
         print 'Arg is of type "bar"'
         ...
      else
         print 'Arg is of a different type'
         ...

这个概念可以根据需要通过不同的方法应用到多个不同的场景中。

Python 3。X包含标准类型库,允许使用@overload装饰器进行方法重载。不幸的是,这是为了使代码更具可读性,因为@overload装饰方法之后需要跟随一个处理不同参数的非装饰方法。 除了你的例子,你可以在这里找到更多:

from typing import overload
from typing import Any, Optional
class A(object):
    @overload
    def stackoverflow(self) -> None:    
        print('first method')
    @overload
    def stackoverflow(self, i: Any) -> None:
        print('second method', i)
    def stackoverflow(self, i: Optional[Any] = None) -> None:
        if not i:
            print('first method')
        else:
            print('second method', i)

ob=A()
ob.stackoverflow(2)

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.