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    <p>You're adding numbers in the wrong "direction". To keep the digits in order, you need to multiply your <code>number</code> variable by 10 and then add <code>array[i]</code> instead of multiplying <code>array[i]</code> and adding it to <code>number</code>.</p> <p>You also need to initialize <code>number</code> to zero before you use it, because a variable has a random value until it's explicitly given one.</p> <p>You also need to do <code>i &lt; 3</code> ("loop while <code>i</code> is less than three") instead of <code>3 &lt; i</code> ("loop while 3 is less than <code>i</code>", which is never true).</p> <pre><code>int array[] = {1,2,3}; int number = 0; for (int i = 0; i &lt; 3; i++) { number *= 10; number += array[i]; } cout &lt;&lt; number; </code></pre> <p>Let's walk through what happens.</p> <ul> <li>In the beginning, <code>number</code> is equal to zero.</li> <li>We get to the loop. First, <code>i</code> equals 0. <ul> <li>We multiply <code>number</code> by 10. For the first iteration, <code>number</code> is still zero after that.</li> <li>We add <code>array[0]</code> (1) to <code>number</code>. <code>Number</code> is now 1.</li> </ul></li> <li><code>i</code> now increments and is equal to 1. 1 is less than 3, so we go in the loop body again. <ul> <li>We multiply <code>number</code> again by 10 to make room for the next digit. <code>number</code> is now equal to 10.</li> <li>We add <code>array[1]</code> (2) to <code>number</code>. <code>Number</code> is now 12.</li> </ul></li> <li><code>i</code> increments and is equal to 2. 2 is less than 3, so we repeat. <ul> <li>We multiply <code>number</code> by 10, again to make room for the next digit. It's now 120.</li> <li>We add <code>array[2]</code> (3) to <code>number</code>, making it 123, the desired result.</li> </ul></li> <li><code>i</code> increments and becomes 3. 3 is obviously not less than 3, so we exit the loop.</li> <li>We print <code>number</code> (123) to the console.</li> </ul>
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    1. COcan you explain what is the meaning of number*= 10; "*" is a pointer right ? and why the value 10?
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    2. CO@user3123562 No, it's not a pointer. `number *= 10;` is equal to `number = number * 10;`
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    3. COYes, `a *= b` is a shorthand for `a = a * b`, the same as `a += b` is a shorthand for `a = a + b`. `*` is *also* used in different contexts for pointers, but that's not the case here.
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