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copied!<p>A 2D array in C is treated as a 1D array whose elements are 1D arrays (the rows).<br> For example, a 4x3 array of <code>T</code> (where "T" is some data type) may be declared by: <code>T a[4][3]</code>, and described by the following scheme: </p> <pre><code> +-----+-----+-----+ a == a[0] ---> | a00 | a01 | a02 | +-----+-----+-----+ +-----+-----+-----+ a[1] ---> | a10 | a11 | a12 | +-----+-----+-----+ +-----+-----+-----+ a[2] ---> | a20 | a21 | a22 | +-----+-----+-----+ +-----+-----+-----+ a[3] ---> | a30 | a31 | a32 | +-----+-----+-----+ </code></pre> <p>Also the array elements are stored in memory row after row.<br> Prepending the <code>T</code> and appending the <code>[3]</code> to <code>a</code> we have an array of <code>3</code> elements of type <code>T</code>. But, the name <code>a[4]</code> is itself an array indicating that there are <code>4</code> elements each being an array of <code>3</code> elements. Hence we have an array of <code>4</code> arrays of <code>3</code> elements each.<br> Now it is clear that <code>a</code> points to the first element (<code>a[0]</code>) of <code>a[4]</code> . On the Other hand <code>&a[0]</code> will give the address of first element (<code>a[0]</code>) of <code>a[4]</code> and <code>&a[0][0]</code> will give the address of <code>0th</code> row (<code>a00 | a01 | a02)</code> of array <code>a[4][3]</code>. <code>&a</code> will give the address of 2D array <code>a[3][4]</code>. <code>*a</code> decays to pointers to <code>a[0][0]</code>.<br> Note that <code>a</code> is not a pointer to <code>a[0][0]</code>; instead it is a pointer to <code>a[0]</code>.<br> Hence </p> <blockquote> <ul> <li>G1: <code>a</code> and <code>&a[0]</code> are equivalent. </li> <li>G2: <code>*a</code>, <code>a[0]</code>and <code>&a[0][0]</code> are equivalent.</li> <li>G3: <code>&a</code> (gives the address of 2D array <code>a[3][4]</code>).<br> But group <strong><em>G1</em>, <em>G2</em> and <em>G3</em> are not identical although they are giving the same result</strong> (and I explained above why it is giving same result).</li> </ul> </blockquote>

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