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Python doesn't have private members, so that doesn't help you as much.

by moonchild 6y ago

heavenlyblue 6y ago
Python has private members through double underscore methods
- ben-schaaf 6y ago
  They're only private in a sense that you're not supposed to access them, there's no enforcement by the language in any way beyond renaming them `_<class>__<attr>`. They're still accessible and they still show up in `dir`, `__dict__`, etc.
- travisjungroth 6y ago
  That’s not their purpose and they’re not really private. Leading double underscore methods cause name mangling (the class name gets added to the method name) and they’re used for solving specific problems with inheritance. Don’t use them just to make something private. You’re screwing up your inheritance if you don’t realize what you’re doing. Leading single underscore is private by convention. Still should generally be avoided, but it’s the proper use.
cessor 6y ago
Members in Python are never private. Variables that are supposed to be used internally only are marked with and underscore, but that just conveys intent and isn't enforced by the interpreter/runtime. But you can emulate private data like this: >>> def a(u, v): ... def b(): ... return u + v ... return b ... >>> b = a(1,2) >>> b() 3 Like this, there is no way to access u & v from b.
- moonchild 6y ago
  Looks like classes can't be closures, though: >>> def f(): ... x = 5 ... class Blub: ... def incx(self): ... x += 1 ... def getx(self): ... return x ... return Blub() ... >>> j = f() >>> j.incx() Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 5, in incx UnboundLocalError: local variable 'x' referenced before assignment
  - cessor 6y ago
    When using it like this, the interpreter opens a new namespace where the variables are bound. The x is declared in the outer scope, so it can't find it. However, you can ask python to keep searching vor the name `x` in the closest outer scope, that is what the `nonlocal` statement is for: def f(): x = 5 class Blub: def incx(self): nonlocal x x += 1 def getx(self): return x return Blub() j = f() j.incx() print(j.getx()) This will print 6, as expected.

Python doesn't have private members, so that doesn't help you as much.

heavenlyblue 6y ago

Python has private members through double underscore methods

ben-schaaf 6y ago
They're only private in a sense that you're not supposed to access them, there's no enforcement by the language in any way beyond renaming them `_<class>__<attr>`. They're still accessible and they still show up in `dir`, `__dict__`, etc.
travisjungroth 6y ago
That’s not their purpose and they’re not really private. Leading double underscore methods cause name mangling (the class name gets added to the method name) and they’re used for solving specific problems with inheritance. Don’t use them just to make something private. You’re screwing up your inheritance if you don’t realize what you’re doing. Leading single underscore is private by convention. Still should generally be avoided, but it’s the proper use.

cessor 6y ago

Members in Python are never private. Variables that are supposed to be used internally only are marked with and underscore, but that just conveys intent and isn't enforced by the interpreter/runtime. But you can emulate private data like this: >>> def a(u, v): ... def b(): ... return u + v ... return b ... >>> b = a(1,2) >>> b() 3 Like this, there is no way to access u & v from b.

moonchild 6y ago
Looks like classes can't be closures, though: >>> def f(): ... x = 5 ... class Blub: ... def incx(self): ... x += 1 ... def getx(self): ... return x ... return Blub() ... >>> j = f() >>> j.incx() Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 5, in incx UnboundLocalError: local variable 'x' referenced before assignment
- cessor 6y ago
  When using it like this, the interpreter opens a new namespace where the variables are bound. The x is declared in the outer scope, so it can't find it. However, you can ask python to keep searching vor the name `x` in the closest outer scope, that is what the `nonlocal` statement is for: def f(): x = 5 class Blub: def incx(self): nonlocal x x += 1 def getx(self): return x return Blub() j = f() j.incx() print(j.getx()) This will print 6, as expected.