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    <p>EDIT: Okay, got it.</p> <p>The first point to make is that obviously you shouldn't use this code anyway. However, when you expand it, it becomes equivalent to:</p> <pre><code>j = j ^ (i = i ^ (j = j ^ i)); </code></pre> <p>(If we were using a more complicated expression such as <code>foo.bar++ ^= i</code>, it would be important that the <code>++</code> was only evaluated once, but here I believe it's simpler.)</p> <p>Now, the order of evaluation of the operands is always left to right, so to start with we get:</p> <pre><code>j = 36 ^ (i = i ^ (j = j ^ i)); </code></pre> <p><strong>This (above) is the most important step.</strong> We've ended up with 36 as the LHS for the XOR operation which is executed last. The LHS is not "the value of <code>j</code> after the RHS has been evaluated".</p> <p>The evaluation of the RHS of the ^ involves the "one level nested" expression, so it becomes:</p> <pre><code>j = 36 ^ (i = 25 ^ (j = j ^ i)); </code></pre> <p>Then looking at the deepest level of nesting, we can substitute both <code>i</code> and <code>j</code>:</p> <pre><code>j = 36 ^ (i = 25 ^ (j = 25 ^ 36)); </code></pre> <p>... which becomes</p> <pre><code>j = 36 ^ (i = 25 ^ (j = 61)); </code></pre> <p>The assignment to <code>j</code> in the RHS occurs first, but the result is then overwritten at the end anyway, so we can ignore that - there are no further evaluations of <code>j</code> before the final assignment:</p> <pre><code>j = 36 ^ (i = 25 ^ 61); </code></pre> <p>This is now equivalent to:</p> <pre><code>i = 25 ^ 61; j = 36 ^ (i = 25 ^ 61); </code></pre> <p>Or:</p> <pre><code>i = 36; j = 36 ^ 36; </code></pre> <p>Which becomes:</p> <pre><code>i = 36; j = 0; </code></pre> <p>I <em>think</em> that's all correct, and it gets to the right answer... apologies to Eric Lippert if some of the details about evaluation order are slightly off :(</p>
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    1. POWhy does swapping values with XOR fail when using this compound form?
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    1. CO@Jon Skeet: Thanks, Now it is clear to me. +100 Great Answer :)
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    2. COSo, in conclusion, this will work: `j = (i ^= j ^= i) ^ j;`, which is about the same, without the leftmost `j`.
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    3. CO@Kobi, yes `j = (i ^= (j ^= i)) ^ j;` is working well.
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