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copied!<pre><code>printf("%d\b",x) </code></pre> <p>prints the characters <code>'1'</code>, <code>'0'</code> (because x==10) and <code>\b</code>. The <code>\b</code> is a backspace character; if you print to a terminal, it will print <code>10</code> and then move the cursor back one column.</p> <p>A call to <code>printf</code> returns the number of characters it printed; in this case, the result is 3 (yes, <code>'\b'</code> counts as a character).</p> <pre><code>printf("%d\n",printf("%d\b",x)); </code></pre> <p>The inner <code>printf</code> call works as I explained above, and returns 3. The outer <code>printf</code> call prints <code>"3\n"</code>.</p> <p>So the entire statement will print:</p> <pre><code>10\b3\n </code></pre> <p>The <code>'\b'</code> causes the <code>3</code> to be replace the <code>0</code> on the screen, so the final displayed result (when I run the program on my system) is:</p> <pre><code>13 </code></pre> <p>If I pipe the output through <code>cat -v</code>, I get:</p> <pre><code>10^H3 </code></pre> <p>where <code>^H</code> represents the backspace character.</p> <p><strong>EDIT :</strong></p> <p>The question was just edited, and the modified program's behavior is quite different. The switch statement causes control to jump <em>past</em> the declaration <code>int x = 10;</code>, but into the scope in which <code>x</code> is declared. As a result, <code>x</code> is uninitialized when <code>printf</code> is called. This causes undefined behavior, and most likely garbage output (I just got <code>-1217572876^H12</code>). If <code>x</code> happens to be <code>0</code>, I suppose you'd get <code>0^H2</code>, which would look like <code>2</code>.</p> <p>Whatever you're trying to do, please find a better way to do it.</p>

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