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    <p>Putting aside the problems with your code, yes, arrays are contiguous elements.</p> <p>What happens with the second line of the code snippet below:</p> <pre><code>int * iptr = malloc (40 * sizeof(int)); int x = iptr[10]; </code></pre> <p>is thus:</p> <ul> <li>the value <code>10</code> is multiplied by the size of the <code>int</code> to get 70 (in this example, we assume a 7-byte <code>int</code>).</li> <li>this is then added to the base pointer <code>iptr</code> (unscaled) and the (7-byte) value is extracted from that location.</li> </ul> <p>This scaling is an important feature. You won't see it when you're working with <code>char</code> arrays but the values <code>&amp;(iptr[0])</code> and <code>&amp;(iptr[1])</code> will be separated by the size of an <code>int</code>, <em>not</em> a <code>char</code> (7 in our example case above).</p> <p>When you add 1 to a pointer, that doesn't actually add 1, it adds the size of the underlying data type. So that:</p> <pre><code>int *p = &amp;(iptr[0]); p++; </code></pre> <p>won't give you an address corresponding to the second <em>byte</em> of that memory, it will actually give you the address of the second <code>int</code> (the seventh byte).</p> <p>To that end, <code>iptr[3]</code> is exactly equivalent to <code>*(i+3)</code> in that they both give you element number three, even though it's made up of <em>bytes</em> from offset 21 through 27 inclusive.</p> <hr> <p>Pre-emptive strike, just in case: keep in mind that a byte in ISO C is <em>not</em> necessarily 8 bits. ISO uses the term <code>octet</code> for that. A <em>byte</em> is the size of a <code>char</code>.</p>
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    1. POWhy does adding to a pointer pointed to the first array member cycle through them in C?
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    1. POWhy does adding to a pointer pointed to the first array member cycle through them in C?
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    1. CO`%p` requires the pointer to be cast to `(void *)` (the usual implicit conversion doesn't happen because it's part of the variable argument list).
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    2. COI am pretty sure that arrays are allocated contiguously. If they weren't, then [] wouldn't work.
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      1. USAlexander Rafferty
    3. COActually you're right, @Alex. I envisaged a scenario where element could be stored at only even physical addresses (leaving odd ones blank) and the C compiler would automatically skip odd ones when doing its calculations. That would work, but I was trying to be too clever by half :-) C99 specifically states "An array type describes a contiguously allocated ..." so, regardless of what would happen at the hardware level, the C "machine" only knows about its own level. Fixed.
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