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    <p><code>*(record + i)</code> is not in fact an address. <code>record</code> is an address, and so is <code>(record + i)</code>, but <code>*(record + i)</code> is the <em>value stored at the address represented by</em> <code>(record + i)</code>. Therefore, calling <code>function(record[i])</code> is the same as <code>function(*(record + i))</code>, which will pass the <em>value</em> of the array element to the function, not a pointer.</p> <p>The syntax <code>&amp;record[i]</code> is not taking the address of an address. It is taking the address of <code>record[i]</code>, which is an object. The braces have a higher precedence than the ampersand, so <code>&amp;record[i]</code> is equivalent to <code>&amp;(record[i])</code>. You can think of it as expanding to <code>&amp;(*(record + i))</code> and then simplifying to <code>(record + i)</code>.</p> <p><strong>Update:</strong> To address your question from the comment, an array "decays" into a pointer if you reference the name of the array by itself. If you add square brackets <code>[]</code>, you will get a <em>value</em> from within the array. So, for your example, say you have an array of structures:</p> <pre><code>struct A { ... char abc[10]; ... } record[10]; </code></pre> <p>Then, you would have:</p> <ul> <li><code>record[i]</code> - an object of type <code>struct A</code> from the <code>record</code> array</li> <li><code>record[i].abc</code> - the <code>abc</code> array inside a particular <code>record</code> object, decayed to a pointer</li> <li><code>record[i].abc[k]</code> - a specific character from the string</li> <li><code>&amp;record[i].abc[0]</code> - one way of creating a pointer to the string</li> </ul> <p>The notation <code>record[i]-&gt;abc</code> that you mention in your comment cannot be used, since <code>record[i]</code> is an object and not a pointer.</p> <p><strong>Update 2:</strong> In regards to your second comment, the same rules described above apply regardless of how you nest the array within a structure (and whether you access that structure directly or through a pointer). Accessing an array using <code>arrayname[index]</code> notation will give you an item from the array. Accessing an array using <code>arrayname</code> notation (that is, using the array name by itself) will give you a pointer to the first element in the array. If you need more details regarding this phenomenon, here are a couple of links that explain arrays and the way that their names can decay into pointers:</p> <ul> <li><a href="http://boredzo.org/pointers/" rel="nofollow noreferrer">http://boredzo.org/pointers/</a></li> <li><a href="http://www.ibiblio.org/pub/languages/fortran/append-c.html" rel="nofollow noreferrer">http://www.ibiblio.org/pub/languages/fortran/append-c.html</a></li> <li><a href="http://c-faq.com/aryptr/index.html" rel="nofollow noreferrer">http://c-faq.com/aryptr/index.html</a></li> </ul>
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    1. POI don't understand one dimensional array handling with function passing
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    1. COsuppose record is a struture with element char abc[10]; inside it now when i do record[i] -> abc it will give address of abc or value inside the string abc
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    2. COnow suppose the record[i] lies inside an structure say having struct type { long rowcount; mytype record[10]; } *mypoint; where mytype is typedef mytype{ char abc[10] } now what result will it give mypoint->record[1].abc ? will it give address or value ? please specify the break down of expression "mypoint->record[1].abc" with regards to record[i] = *(record + i)
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    3. COIf you want to extend your question, edit your question and add this information above. Posting it in a comment is both harder to read and easier to ignore. Also, is this homework? The questions seem to be worded as such.
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