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    <p>Okay, let's review first, how arrays decay and second, how declarations work.</p> <p><code>A</code> is declared as a 2D array. When <code>A</code> is used in any expression, it will "decay" to a pointer to its first element. In C, a 2D array is made up of two 1D arrays. So an element of a 2D array is a 1D array. So when <code>A</code> is used in any expression, it will decay to a pointer to the first row of <code>A</code>, which is an array of 10 ints. </p> <p><code>int (*v) [10]</code> means that <code>v</code> is a pointer to an array of 10 ints. The assignment</p> <pre><code>int (*v)[10] = A; </code></pre> <p>is an expression, so <code>A</code> there has decayed to a pointer to 10 ints as well, so the assignment is valid. </p> <p>So now when you look at <code>**v</code>, you are dereferencing first, <code>v</code>, which means you are getting what <code>v</code> points at, which is the first row of your 2D array. Then you are dereferencing <code>*v</code>, which means you are getting the first element of that first row. That is 1, so your output is 1.</p> <p>When you look at <code>**(v + 1)</code>, you are first adding 1 to the pointer <code>v</code>. Since <code>v</code> points to a row, adding 1 gets you a pointer to the next row. Then you do the double dereferencing as above and you get the first element of the next row, which is 11.</p> <p>When you look at <code>*(*v + 1)</code>, you are first dereferencing <code>v</code>, which means you get the first row. Then you add 1 to that, which gets you a pointer to the next element of the row. Then you dereference that, which gets you the 2, the second element. </p> <p>To summarize: <code>v</code> points to the whole first row. <code>*v</code> points to the first element of the first row. <code>(v + 1)</code> points to the whole second row. <code>(*v + 1)</code> points to the second element of the first row. </p> <p>After this, figuring out the rest is probably quite easy. </p>
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    1. POAbout C: pointer and Two-dimensional array
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    1. COThank you very much, I think here assigning 10 elements to v makes me confused. In fact, int (*v)[2] is enough.
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    2. CONo, it won't be enough. `(*v)[2]` would be a pointer to an array of 2 ints. Each row in `A` is an array of 10 ints. You really need `(*v)[10]`.
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