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<pre><code>A d = new A().aob = new A(); </code></pre> In Java, the assignment operator is right-associative, i.e. they are evaluated from right-to-left. But also, they are in the least precedence operators group. So the second new operator (right of the second equality) is evaluated first and we get a new A object; let's say 'a'. Now we have: <pre><code>new A().aob = a; </code></pre> The trick here is to recognize the operator precedence. Take a look at here: <a href="http://pages.cs.wisc.edu/~willb/cs302/spring-07/java-operator-precedence.pdf" rel="nofollow">http://pages.cs.wisc.edu/~willb/cs302/spring-07/java-operator-precedence.pdf</a> The 'new' operator and the '.' method call operator have the same precedence but their associative quality are reversed: 'new' is right-associative and '.' is left-associative. So the compiler first applies the new operator on the 'right operand' which is here 'A()' (before the next operand comes into place). Let's call the new object b; and we have: <pre><code>A d = b.aob = a; </code></pre> The compiler now needs to apply the '.' operator first (as '.' has higher precedence than '=' operator). Let's call the object referred by 'b.aob' as c: <pre><code>A d = c = a; </code></pre> Finally all that is left are the assignment operators and they are right-associative (evaluated from right-to-left). So, a is assigned to c (b.aob) first, and then c is assigned to d.
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