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2013-08-17T12:41:05.380
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What gave you the wrong output in the first case was the fact that the pointer <code>Q->currentValue</code> never changed its value (the address which it was holding). It was always pointing to the first element in the queue. Say the queue contained <code>{1, 2, 3 |,<garbage>}</code>. That means, that after the first pop, the queue became: <code>{2, 3 |, 3, <garbage>}</code> and <code>currentValue</code> still held the address of the first element in the array, which is <code>2</code>. After the second pop: <code>{3 |, 3, 3, <garbage>}</code> and <code>currentValue</code> points to the first element, which value is <code>3</code>, The last time, the array is unchanged (as <code>Q->count--</code> changes <code>Q->count</code>'s value to 0), so the conent is <code>{| 3, 3, 3, <garbage>}</code> and <code>currentValue</code> is still pointing to a <code>3</code>. I assume that you changed the second example to make <code>Queue->currentValue</code> an <code>int</code>. This way, it retains the original first element (which is popped). That makes your prints work properly in your test case. However, <ul> <li>Your implementation will fail if you had a <code>0</code> in your queue.</li> <li>Your implementation adds needless complexity to the <code>pop</code> operation (O(n)). It would be much better to implement a cyclic queue, with <code>head</code> and <code>tail</code>.</li> <li>Retaining a copy of the popped element in order to return it would not be my first choice. I would suggest implementing an <code>isEmpty()</code> method, check its result in the <code>while</code> loop and while the queue is not empty, simply have <code>pop()</code> return the queue advance its <code>head</code> and return the previous head element.</li> </ul>
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1. COthank you very much @MasterAM for the clarification and for the advice's .. god bless you
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2. COYou are welcome. If this answers your question, the next thing to do is accept the answer :)
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