6 ms·

That seems like decidedly unexpected behaviour and makes default params far less useful.

• jsprinkles 14y ago
  Actually, if you understand the way Python is evaluated (dig in, the core is pretty transparent), it's the only behavior that makes sense in this case. It's also documented as such[1], so it's quite expected. Default parameters are still just as useful for constants, such as: def f(x=0, y="foo", z=3.14159): This, however, is a perfectly Pythonic idiom: def f(L=None): if L is None: L = [] [1]: http://docs.python.org/reference/compound_stmts.html#function-definitions http://docs.python.org/reference/compound_stmts.html#functio...
  • swdunlop 14y ago
    Or just embrace the tao of the tuple: def f(L=()): ...
  • ollysb 14y ago
    While it seems logical when you understand what's going on, from a practical point of view I can't see how this would ever be useful. The tradeoff appears to be that the functions are first class objects. I'm not sure what the benefit here is though. Does having them as first class objects allow some useful idioms? (I'm a ruby dev but I'm genuinely curious to know what this allows you to do)
    • jsprinkles 14y ago
      It's not designed to be useful, it just is. Functions are objects, yes, and the def statement brings them into being and assigns them to the given name in the current scope: >>> def f(x): ... return x + 5 >>> type(f) <type 'function'> >>> dis.dis(f.func_code) 2 0 LOAD_FAST 0 (x) 3 LOAD_CONST 1 (5) 6 BINARY_ADD 7 RETURN_VALUE >>> g = f >>> g(10) 15 >>> g is f True They're just variables in the current scope. If you're quite clever, your brain is already figuring out that has some interesting implications that some libraries use: >>> import socket >>> socket.gethostbyname('www.google.com') '74.125.71.103' >>> socket.gethostbyname = lambda i: '10.0.0.1' >>> socket.gethostbyname('www.google.com') '10.0.0.1' I will not pass judgement on monkey patching like this, just pointing out it's doable. I know for a fact Ruby can as well. Functions just being variables has useful properties when you're doing something like fancy switch/case type things (the readability of this is questionable, but it's cool to look at it, like a Duff's device): >>> i = 1 >>> { str: func1, unicode: func2, int: func3 }.get(type(i), func1)(i) in func3 Also consider something like this, which is how decorators work (and they're incredibly useful), sort of like a closure: >>> def maker(i): ... def ret(x): ... return x + i ... return ret >>> f, g = maker(10), maker(100) >>> f(5), g(10) (15, 110) I'd be surprised if Ruby couldn't do everything I just did.
      • mattmanser 14y ago
        It's not designed to be useful By useful he means mucking up your program in totally unexpected ways. He's being nice about it being a silly decision to have it behave that way. The entire post reads more like a list of unexpected things that will bite you in the ass.
      • adi92 14y ago
        Thanking you for showing me the 'dis' module. Dis is going to be fun to play with!!
    • philwelch 14y ago
      Functions (methods) are first class objects also in Ruby: methods are instances of the class Method, while Procs are a lightweight alternative. Default arguments in Ruby are not mutable in any kind of function object, be it a lambda proc, a regular proc, or a method. You can use a mutable default argument as an ersatz static variable, e.g. for memoization.
      • bodhi 14y ago
        > Default arguments in Ruby are not mutable in any kind of function object Maybe I misunderstood you, but they are perfectly mutable: class A attr_accessor :a def initialize @a = [] end def b x = a x << 1 end end obj = A.new # => #<A:0x007fd92b2e9850 @a=[]> obj.b # => [1] obj.b # => [1, 1] obj.a # => [1, 1] If you mean "inline default arguments are not mutable", that's not true either. What is true is that the default argument is evaluated when the function is called, not when it is defined: a = 0 # => 0 x = lambda {|y = (a + 1)| y } # => #<Proc:0x007fd92b1dae50@(irb):33 (lambda)> x[] # => 1 a = 5 # => 5 x[] # => 6
        • philwelch 14y ago
          Good catch, I hadn't thought of the case where the default argument is expressed in terms of another variable.
        • mratzloff 14y ago
          And this is expected behavior. What is unexpected is when this behavior is afforded to new list or dictionary arguments in Python, just as it would be unexpected to say ruby> class A; end => nil ruby> def foo(bar = A.new); return bar; end => nil ruby> foo => #<A:0x00000101985060> ruby> foo => #<A:0x0000010197a6b0> and get back the same object each time `foo` is called in Ruby. I've even seen a major Python library with this bug (I'm sorry, I don't recall which off-hand). It's really surprising behavior for new Python devs.
    • anthonyb 14y ago
      There are a few use cases for default variables on effbot's site: http://effbot.org/zone/default-values.htm http://effbot.org/zone/default-values.htm Basically, sometimes you do want to reuse the mutable between function calls, and in those cases it can save a fair bit of code passing it in repeatedly.
      • iki23 14y ago
        Good coverage. I use it quite often for cache dictionary, it's much simpler API and overall code, than creatng a new class for it. Demo snippet from effbot's site: def calculate(a, b, c, memo={}): try: value = memo[a, b, c] # return already calculated value except KeyError: value = heavy_calculation(a, b, c) memo[a, b, c] = value # update the memo dictionary return value
        • ollysb 14y ago
          This seems a bit leaky. You're exposing the caching mechanism in the method signature(yeah, ok, in practice it's unlikely to be a problem).
          • anthonyb 14y ago
            The other option is to create your own cache object and pass that in (and around, if it's a recursive function). Of course, in Python pretty much every cache object follows the dictionary interface anyway, so it doesn't really matter. One of the benefits of duck typing :)
  • chernevik 14y ago
    I am but an egg, but isn't this the same but shorter? def f(L=None): L = L or []
    • coderdude 14y ago
      It might work the same depending on your use of it but it's not the same. In that instance L will become a blank list if it is equal to None, zero, or a zero length string. There are many cases where this wouldn't affect anything, but there can also be instances where that will cause you to define L as a blank list when you really wanted to keep L's value. I think it's always better to be explicit and test for the value(s) you expect.
      • chernevik 14y ago
        Ah. These are the little assumptions that keep blowing off my feet. Thanks.
        • SoftwareMaven 14y ago
          It is these cases that brought the ternary operator to Python: def f(x=None): x if x is not None else []
          • iki23 14y ago
            Yes, but why? Actually, it's No, because python culture aims to use one way to do thing, the least surprising one. In this case it's: def f(x=None): if x is None: x = []
            • SoftwareMaven 14y ago
              Because that "if" is a statement whereas the ternary expression is, well, an expression. There are places expressions can be used that statements can't (eg lambdas) and that x or [] won't work (eg when x is False). That said, people still seem to favor your form as the more Pythonic way. Personally, I think that's just because the ternary expression is relatively new.
• Estragon 14y ago
  It's like whitespace. Everybody has this reaction at first, then they get over it.
  • gbog 14y ago
    Ahem, we have repeatedly intricate issues that are related to this kind of behavior, and I have yet to see an annoying white space bug. Not to say I think python should be changed on this point. It shouldn't, there are code checkers that warn you on the gotcha, let's use them.
• makecheck 14y ago
  The thing is, None is always a possible value for a parameter so it's actually more robust if functions are written to expect None. If you say "f(x=[])" (assuming that worked without the actual side effects it has), someone could still say "x(None)" instead of "x()", causing the function to die. Since a robust program isn't able to avoid checking for None, it might as well set defaults there too. There is another case where this is important; you might want the equivalent of "f(x=expensive_function_to_calculate_useful_default())", and you don't want that function called unless it needs to be. Only the x=None approach allows this to be deferred.
  • anthonyb 14y ago
    I disagree. If you want your programs to be robust like that, you now have to check for every case where someone might pass in something stupid (dict instead of int, maybe?). Much better to catch errors further up the chain and keep your low level code simple (ie. pass me something other than an iterable and I blow up). In the expensive case, I'd just calculate it once and store it somewhere (possibly as a lookup dictionary if there are multiple inputs) and access that from within the function.
    • makecheck 14y ago
      Well that's true, a function is generally written as if it's been given what it wants. I wouldn't check for other types either. But None is a result that can happen in situations that would otherwise return exactly the expected type. If "nothingness" can be meaningful (especially in a function that accepts an empty list as a parameter, say), it's nicer if the code just deals with None itself instead of requiring checks for None in all the callers.
• ScottBurson 14y ago
  It's a mistake. Default values for optional parameters don't act that way in any other language I know of, including Common Lisp, in which functions are also firstclass.
  • jsprinkles 14y ago
    It's most certainly not a mistake; Python 3 would probably have fixed it, if it were. It is an (admittedly, strange) side effect of the way 'def' works.
    • ScottBurson 14y ago
      But 'def' doesn't have to work that way. Consider, in CL: > (defvar *fn* (let ((x 3)) (lambda (&optional (y (list nil x))) (push 7 (car y)) ; modifies the list y))) *FN* > (funcall *fn*) ((7) 3) > (funcall *fn*) ((7) 3) From this example you can see two things. First, the binding of 'x' is closed over when the lambda expression is evaluated. And second, the expression that provides the default value of 'y' is evaluated every time the function is called. There's no fundamental reason it couldn't have worked that way in Python. (I understand that changing the language so it worked that way now would likely break some code.) EDIT: fixed formatting.
      • Estragon 14y ago
        Expensive, though. http://stackoverflow.com/questions/1651154/why-are-default-arguments-evaluated-at-definition-time-in-python http://stackoverflow.com/questions/1651154/why-are-default-a...
        • ScottBurson 14y ago
          Surely not. The implementation already has to check that the number of provided arguments is valid. The decision of whether to evaluate the default expression can be part of that. The code that evaluates the default expression doesn't need to be in a separate function, either, so the argument that calling that function is too expensive also doesn't hold water. I just tried a test in SBCL: (defun foo1 (x) x) (defun test1 (n) (dotimes (i n) (foo1 (cons nil nil)))) (time (test1 100000000)) => 4.4 sec, or 44ns / iteration (defun foo2 (&optional (x (cons nil nil))) x) (defun test2 (n) (dotimes (i n) (foo2))) (time (test2 100000000)) => 4.1 sec, or 41ns / iteration The version with the optional parameter is actually slightly faster, which completely blows a hole in the performance argument. Look, no language is perfect -- not even Common Lisp :-) I think users are better served when design flaws in a language are acknowledged without defensiveness than when bogus justifications are offered.