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    <p>A detail here, the "inequalities" you're asking about are about pointers, not arrays. Think of comparing memory locations, and the following will be easier to follow.</p> <ul> <li>First one should be pretty straightforward: print each element of the array in order (<code>a[0]</code>, <code>a[1]</code>, etc.)</li> <li>In number 2, <code>p</code> is the address of element <code>0</code> of <code>a</code>. <code>p++</code> increases that address to be address of the next element in the array.The termination condition can be read as "<code>p</code> is less or equal the address of element number 4 of the array". To put some values and make this concrete: imagine <code>&amp;a[0]</code> is <code>200</code>, as a consequence <code>&amp;a[4]</code> is <code>216</code>, as <code>p</code> is incremented it becomes <code>204, 208, 212, 216</code>, and finally <code>220</code>. When it becomes <code>220</code> the condition becomes false, and the loop ends.</li> <li>Number 3's left side is similar but instead of increasing <code>p</code> it adds <code>i</code> to <code>p</code>. Note that thanks to the magic of pointer arithmetic <code>p + 1</code> is actually <code>p + i*(sizeof(*p))</code>. The sequence is exactly the same as in case 2. BTW, initialization should be <code>p = &amp;a[0]</code>, <code>i=0;</code> (note the comma before the <code>i</code>). For the right side <code>(a+4)</code>, the plain a means the address of the first element of a (this is the C equivalency between pointers and arrays, <code>a</code> topic full of subtleties and pain). <code>a + 4</code> is thus equivalent to <code>&amp;a[4]</code>.</li> <li>Number 4 is there to mess with your head. The information above should be enough to translate the initialization and condition; but having both <code>i++</code> and <code>p++</code> means you will be advancing 2 elements each iteration.</li> </ul> <p>The use of the preprocessor here is likely obscuring what gets called, try expanding manually to follow up more easily. and PRD should be adding a space after each number for easier reading.</p>
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