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.
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.
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:
while method call is defined in:
and both are used here:
so what happens is that
is parsed as
So there is no choice according to priority because the
Primary
is reduced before. Mind that for the special situation likethe 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.