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0 pointsKranar3y ago0 comments

No sorry, you are incorrect and you can verify the correct answer on Wikipedia or any technical book on this subject.

For example the Java Language Specification states in section 8.4.1 that Java is pass by value:

https://docs.oracle.com/javase/specs/jls/se8/html/jls-8.html...

ECMAScript's Language Specification also states in section 10.6 that it uses pass by value semantics, although it's much more formal about the specific approach it uses:

https://262.ecma-international.org/5.1/

I can't link to the specific section but you can review the semantics of MakeArgGetter and MakeArgSetter which are specified to produce arguments bound to the *value* associated with the name, as opposed to a reference.

Python does not have a spec that I can reference, but given that its argument semantics follows those of Java, and once again the inability to write a swap function, it should not be too difficult to deduce that Python also passes by value.

0 comments

rockemsockem3y ago

This is indeed correct. Object references themselves in Java/Python are passed by value, which is why the following two code samples do not have the same effect:

  Python 3.8.10 (default, Jun 22 2022, 20:18:18) 
  [GCC 9.4.0] on linux
  Type "help", "copyright", "credits" or "license" for more information.
  >>> def f(d):
  ...     d['foo'] 'bar'}
  ... 
  >>> a = {}
  >>> f(a)
  >>> a
  {'foo': 'bar'}



  Python 3.8.10 (default, Jun 22 2022, 20:18:18) 
  [GCC 9.4.0] on linux
  Type "help", "copyright", "credits" or "license" for more information.
  >>> def f(d):
  ...     d = {'foo': 'bar'}
  ... 
  >>> a = {}
  >>> f(a)
  >>> a
  {}

2 more replies

icedchai3y ago

I said they pass objects by reference, which is true. Technically, yes, Java is pass by value. However, objects are not passed by value. The value you are passing is a reference (pointer) to that object. If this were not the case, you would not be able to see modifications to method parameters reflected in the caller because you're modifying a copy. You are not modifying a copy.

KranarOP3y ago

I think you are deeply confused about this topic and should not be making judgements about other people. Instead of making an honest attempt to understand the difference between pass by value and pass by reference, you are trying to mince words to save face and even your attempt to mince words is incorrect.

While I likely can not convince you further that you are in error, especially since you're now trying to double down on this, for others reading this who have a genuine desire to understand this topic, please understand that Java, JavaScript and Python do not pass by reference, but instead pass by value and refrain from attempting to redefine technical terminology.

The references I cite are quite authoritative and unambiguous on this topic.

icedchai3y ago

I already agreed you were technically correct!

j / k navigate · click thread line to collapse

0 pointsKranar3y ago0 comments

No sorry, you are incorrect and you can verify the correct answer on Wikipedia or any technical book on this subject.

For example the Java Language Specification states in section 8.4.1 that Java is pass by value:

https://docs.oracle.com/javase/specs/jls/se8/html/jls-8.html...

ECMAScript's Language Specification also states in section 10.6 that it uses pass by value semantics, although it's much more formal about the specific approach it uses:

https://262.ecma-international.org/5.1/

0 comments

rockemsockem3y ago

This is indeed correct. Object references themselves in Java/Python are passed by value, which is why the following two code samples do not have the same effect:

  Python 3.8.10 (default, Jun 22 2022, 20:18:18) 
  [GCC 9.4.0] on linux
  Type "help", "copyright", "credits" or "license" for more information.
  >>> def f(d):
  ...     d['foo'] 'bar'}
  ... 
  >>> a = {}
  >>> f(a)
  >>> a
  {'foo': 'bar'}



  Python 3.8.10 (default, Jun 22 2022, 20:18:18) 
  [GCC 9.4.0] on linux
  Type "help", "copyright", "credits" or "license" for more information.
  >>> def f(d):
  ...     d = {'foo': 'bar'}
  ... 
  >>> a = {}
  >>> f(a)
  >>> a
  {}

2 more replies

icedchai3y ago

KranarOP3y ago

The references I cite are quite authoritative and unambiguous on this topic.

icedchai3y ago

I already agreed you were technically correct!

j / k navigate · click thread line to collapse