Java casting order

by

(C)a.foo() is equivalent to (C)(a.foo()), i.e. #2 in the question.

To get #1, you would have to write ((C)a).foo().

The Java language specification does not specify operator precedence in a nice, easy-to-read summary.

Appendix A of Introduction to Programming in Java by Sedgewick and Wayne has a comprehensive table of operator precedence.

Appendix B of The Java Programming Language has a table of operator precedence, but it is not as complete as Sedgewick’s.

A close inspection of the grammar in the Java Language Specification can determine the relative precedences of the cast and method call expressions in question:

Expression:
        Expression1 [AssignmentOperator Expression1]]

Expression1:
        Expression2 [Expression1Rest]

Expression1Rest:
        ?   Expression   :   Expression1

Expression2 :
        Expression3 [Expression2Rest]

Expression2Rest:
        {InfixOp Expression3}
        Expression3 instanceof Type

Expression3:
        PrefixOp Expression3
        (   Expression | Type   )   Expression3
        Primary {Selector} {PostfixOp}

Primary:
        ParExpression
        NonWildcardTypeArguments (ExplicitGenericInvocationSuffix | this Arguments)
        this [Arguments]
        super SuperSuffix
        Literal
        new Creator
        Identifier { . Identifier }[ IdentifierSuffix]
        BasicType {[]} .class
        void.class

The relevant productions are bolded. We can see that a cast expression matches the production Expression3 : (Expression|Type) Expression3. The method call matches the production Expression3 : Primary {Selector} {PostfixOp} by means of the production Primary: Identifier {. Identifier }[IdentifierSuffix]. Putting this together, we see that the method call expression will be treated as a unit (an Expression3) to be acted upon by the cast.

Hmmm, the precedence chart is easier to follow… 😉

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