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    <p>The expression <code>x[y]</code> is equivalent to <code>*(x+y)</code> hence it makes no difference whether you state it as <code>x[y]</code> or <code>y[x]</code>.</p> <p>In fact, <code>"0123456789abcdef"[nybble % 0x10]</code> is also a valid expression, returning one of those characters depending on the value within the brackets. So is the alternative, as shown in this program:</p> <pre><code>#include &lt;stdio.h&gt; int main() { char *arr = "0123456789abcdef"; int nybble = 11; printf ("%c%c\n", arr[11], 11[arr]); return 0; } </code></pre> <p>which prints <code>b</code> twice. So, no, <code>c</code> doesn't have to be an array at all.</p> <p>In terms of why it's always true, just convert them to the <code>*(x+y)</code> form:</p> <pre><code> a[b[c]] == c[b][a] =&gt; a[*(b + c)] == (*(c + b))[a]) =&gt; *(a + *(b + c)) == *(*(c + b) + a) </code></pre> <p>Now, because <code>a + *(b + c)</code> is mathematically exactly the same as <code>*(c + b) + a</code>, both sides of the equation reference the same memory location, hence the expression will always be true.</p> <hr> <p>Now keep in mind that the compiler <em>may</em> be smart enough to figure out that it's the same thing and optimise it to always give you true, but you're still straying into undefined behaviour territory.</p> <p>Because you don't explicitly set <code>c</code> to a value, and it's not of static storage duration, it will be given an arbitrary value. That means that you may actually use an idex into the <code>b</code> array that's beyond the end of the actual array, a big no-no in C circles.</p>
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    1. COBut remember: The standard forbids to access an array declared as `arr[5][5]` by `arr[6][0]` as the standard says, you are not allowed to break out of the array boundaries which are 5 and 5 (or as index 4 and 4) in my example ;)
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    2. CO@Zaibis, you caught me mid-edit, I've explained why it's a bad idea and undefined behaviour.
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    3. COThanks for clarifying
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