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Calling a method on a new object in Java without parentheses: order of operations violation? [closed]

According to this table of Java operator precedence and associativity, member access has higher precedence than the new operator.

However, given a class myClass and a non-static member function myFunction, the following line of code is valid:

new myClass().myFunction();

If . is evaluated before new, how can this line be executed? In other words, why aren't parentheses required?

(new myClass()).myFunction();

My guess is that since () shares precedence with ., the myClass() is evaluated first, and so the compiler knows even before evaluating the new keyword that the myClass constructor with zero parameters is being called. However, this still seems to imply that the first line should be identical to new (myClass().myFunction());, which is not the case.

like image 979
Kyle Strand Avatar asked May 12 '13 23:05

Kyle Strand


2 Answers

This is because of how the grammar of Java language is defined. Precedence of operators comes into play just when the same lexical sequence could be parsed in two different ways but this is not the case.

Why?

Because the allocation is defined in:

Primary: 
  ...
  new Creator

while method call is defined in:

Selector:
  . Identifier [Arguments]
  ...

and both are used here:

Expression3: 
  ...
  Primary { Selector } { PostfixOp }

so what happens is that

new myClass().myFunction();

is parsed as

         Expression
             |
             |
    ---------+--------
    |                |
    |                |
  Primary        Selector
    |                |
    |                |
 ---+---            ...
 |     |
new   Creator 

So there is no choice according to priority because the Primary is reduced before. Mind that for the special situation like

new OuterClass.InnerClass()

the class name is actually parsed before the new operator and there are rules to handle that case indeed. Check the grammar if you like to see them.

like image 82
Jack Avatar answered Sep 26 '22 18:09

Jack


I disagree with the conclusion drawn from Jack's diagram. When a grammar is written its nonterminals and structure are designed to implement the precedence and associativity of the language being described. That's why the classic BNF for expressions introduces the "term" and "factor" nonterminals - to enforce the normal multiplication before addition precedence of arithmetic.

So the fact that "Primary -> new Creator" and "Expression -> Primary Selector" in the grammar means that "new Creator" is at a higher precedence level than "Primary Selector".

It seems to me that the grammar is evidence that the Java operator precedence and associativity table is incorrect.

like image 33
R. Mongiovi Avatar answered Sep 22 '22 18:09

R. Mongiovi