(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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