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    <p>No braces means that the loop affects only the next <strong>statement</strong> that follows the loop.</p> <p>So </p> <pre><code>for (i = 2; i &lt;= n; i = i+2) System.out.println (n + " " + i + " " + j ); </code></pre> <p>is equivalent to </p> <pre><code>for (i = 2; i &lt;= n; i = i+2) { System.out.println (n + " " + i + " " + j ); } </code></pre> <p>Usually, indentation is used in those cases to make the code more comprehensible, like this:</p> <pre><code>for (i = 2; i &lt;= n; i = i+2) System.out.println (n + " " + i + " " + j ); </code></pre> <p><strong>EDIT:</strong> Now, this is the actual answer to the question. It all depends on the different iterations the loop does and how do the variables get incremented.</p> <pre><code>int n = 8; int i = 1; int j = 1; j = n + 2; //This means that j takes the value 10. System.out.println (n + " " + i + " " + j ); // 8 1 10 So far, so good. </code></pre> <p>Now, on with the iteration:</p> <pre><code>while (n &gt; 1) { n = n/2; for (i = 2; i &lt;= n; i = i+2) System.out.println (n + " " + i + " " + j ); j++; } </code></pre> <p>For the first iteration, we have <code>n=8 i=1 j=10</code>, so since <code>n &gt; 0</code> is <code>true</code> the loop takes place.</p> <pre><code>n = n / 2; //n = 4 </code></pre> <p>Then, the <code>for</code> (note that it just assigns the value <code>2</code> to <code>i</code>):</p> <pre><code>for (i = 2; i &lt;= 4; i = i+2) //Since n = 4 </code></pre> <p>Since <code>n = 4</code>, the only values that <code>i</code> can take are <code>2</code> and <code>4</code>, then the prints are:</p> <pre><code>4 2 10 4 4 10 </code></pre> <p>After that, <code>j</code> is incremented by one, making it <code>j = 11</code>. The condition for the <code>while</code> is met again because <code>n = 4</code>. <code>n = n/2</code> makes <code>n</code> take the value <code>2</code>, so it enters the <code>while again</code>. Let's take a look at the for again:</p> <pre><code>for (i = 2; i &lt;= 2; i = i+2) //Since n = 2 </code></pre> <p>This time, the only value that <code>i</code> can take is <code>2</code> (note that the value of <code>i</code> is reset again to <code>2</code> while starting the iteration), and that's the print you get.</p> <pre><code>2 2 11 </code></pre> <p>Before iterating again, <code>j++</code> makes <code>j</code> have the value <code>12</code>.</p> <p>On the final iteration, <code>n = n/2</code> results in <code>n = 1</code> since n is an <code>int</code> but this is done inside the while, so it enters again. Now that <code>n = 1</code> the loop looks like this:</p> <pre><code>for (i = 2; i &lt;= 1; i = i+2) //Since n = 1 </code></pre> <p><code>i</code> is set to <code>2</code> and the condition for the for is not met (<code>2 &lt;= 1</code> is <code>false</code>). Then there is no print this time, yet <code>j</code> is incremented to <code>13</code> at the end of the while.</p> <p>In the next iteration you have <code>n = 1</code>, that means that the <code>while</code>'s condition is not met so the iteration breaks. Finally you have <code>n = 1</code>, <code>i = 2</code> and <code>j = 13</code>. That's the last print you get.</p>
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    1. POHow do I read a loop that has no braces?
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    1. CO"The following **statement** of code". You could have many statements in one line.
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    2. COThanks @RobertHarvey for the edit, I was actually adding the same thing.
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    3. COThe OP stated in the question that they already know this fact. It's likely something else that they're missing.
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