3 ms·

Author here. Questions/comments/feedback are very much appreciated! AMA

• re 10y ago
  Very minor point: be careful comparing IDs the way you do in the REPL (i.e., without keeping a reference to the previous object)--since the old one is released, it's possible that the memory address is reused and the ID of the new object ends up being the same, even though the object is different, which can have misleading results: Python 2.7.10 (default, Oct 23 2015, 19:19:21) [GCC 4.2.1 Compatible Apple LLVM 7.0.0 (clang-700.0.59.5)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> class A: ... def b(self): pass ... >>> A.b <unbound method A.b> >>> hex(id(A.b)) '0x104e2c0a0' >>> hex(id(A.b)) '0x104daeb90' >>> hex(id(A.b)) '0x104daeb90' So the `x is y` approach is "safer." :)
  • avyfain 10y ago
    Interesting, had not considered that at all. So basically, to ensure the old reference is not released, just naming it would do the trick? AKA: >>> some_unbound_method = A.b >>> hex(id(some_unbound_method)) '0x104e2c0a0'
    • phasmantistes 10y ago
      Yep, that's right. To lay out the example the way you did in the article, it would look like: >>> foo = A.b >>> bar = A.b >>> hex(id(foo)) '0xdeadbeef' >>> hex(id(bar)) '0xdecafcab' >>> foo is bar False