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copied!<p>I'm glad you asked, because few people at the other question attempted to explain <em>why</em> it was that way (<a href="https://stackoverflow.com/questions/3831341/why-does-this-go-into-an-infinite-loop/3832184#3832184">plodoc</a>'s answer was probably closest). The most important part is:</p> <p><a href="http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.7.2" rel="nofollow noreferrer">§15.7.2</a> Evaluate Operands before Operation:</p> <blockquote> <p>"The Java programming language also guarantees that every operand of an operator (except the conditional operators &&, ||, and ? :) appears to be fully evaluated before any part of the operation itself is performed."</p> </blockquote> <p>The <code>=</code> here is the Simple Assignment Operator = (<a href="http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.26.1" rel="nofollow noreferrer">§15.26.1</a>)</p> <p>It also guarantees the left operand will be evaluated first (<a href="http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.7.1" rel="nofollow noreferrer">§15.7.1</a>), but the left hand side is trivial in this case.</p> <p>This means the <code>x++</code> must be evaluated before the assignment operator is evaluated. The increment will take place before the assignment, and since it's post-increment, it evaluates to <code>x</code>'s old value.</p> <p>Finally, "old x" will be stored in x when the assignment operator is evaluated.</p> <p>It is important to emphasize that this is not guaranteed in all languages regardless of what it does in your compiler, or how intuitive it seems (or not).</p> <p>It's also <em>not</em> a question of precedence. <code>++</code> has higher precedence than <code>=</code> in C and C++ too. But in those languages, modifying a variable twice between sequence points is undefined behavior. So fully compliant C compilers yield different results. </p>

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