Java 'new' Operator Precedence

Although many sources place the new operator on the same precedence level as the unary operators, it is, in fact rather higher. This common error is probably a result of copying from C/C++ precedence tables.

If you would like to verify this yourself, take a look at the Java Language Specification. However, Java coders usually know this already when they code things like new SomeClass().memberAccess

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