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    <p>Your question is probably best answered by an explanation of what <code>void</code> means.</p> <p><code>void</code> basically denotes the absence of a type. As a consequence of this, you cannot tell the compiler to create an object of type <code>void</code>, e.g. with the following statements:</p> <ul> <li><code>new void</code></li> <li><code>new void[...]</code></li> </ul> <p>The meaning of these expressions is roughly the following: "create something", or "create an array of somethings" respectively. This is clearly not a precise enough statement for the compiler.</p> <p><code>void*</code> (pointer to <code>void</code>) however is permitted because pointers are fundamentally always the same thing to the compiler: a memory address of another object. You can think of a pointer as some kind of arrow pointing to something. If you're working with a pointer, the compiler sees that arrow, and not the actual thing it points at. Therefore the compiler doesn't care that the "target type" is <code>void</code>.</p> <p>The consequence of this, however, is that you cannot dereference a <code>void*</code> (pointer to <code>void</code>), because then you make the compiler effectively look at the pointed-to thing, which would be a <code>void</code> value, which doesn't make any sense to the compiler.</p> <p><strong>Summary</strong>:</p> <p><strong>1)</strong> You <em>cannot</em> create an array of type <code>void</code>, as in <code>new void[...]</code>.</p> <p><strong>2)</strong> You <em>can</em> create a pointer to <code>void</code> (<code>void*</code>), or even a pointer to a pointer to <code>void*</code> (<code>void**</code>). </p> <p><strong>3)</strong> You <em>cannot</em> dereference a <code>void*</code> pointer (but you <em>can</em> dereference a <code>void**</code> pointer).</p> <p><strong>Conclusions</strong>:</p> <p><strong>4)</strong> You <em>can</em> create an <code>int*</code> array and let a <code>void*</code> refer to it:</p> <pre><code>int** m; // ... (create the dangling array as in the OP's code and let m point to it) void* v = (void*)m; </code></pre> <p>(See the comments below on why a <code>void*</code> is used here instead of <code>void**</code>!)</p> <p><strong>5)</strong> Because of statement #3, all you can reasonably do with such a pointer is pass it around, but you <em>cannot</em> work on the array's actual content. In order to do this, you need to type-cast it back to the correct data type:</p> <pre><code>int **m2 = (int**)v; </code></pre>
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    1. POHow do you use a void pointer to generate 2D dynamic Array?
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    1. COIn `void **v = (void **)m;`, `v` is not a `void` pointer. Did you mean `void *v = (void *)m;`?
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      1. USAlok Singhal
    2. COTrue, statement #4 and the code example don't fit together nicely. I'll correct this.
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    3. COIt's still weird. Why do you write `void** v = (void**)m;` instead of `void* v = m;`? (notice http://stackoverflow.com/questions/560845/what-are-the-often-misunderstood-concepts-in-c/566416#566416 )
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      1. USJohannes Schaub - litb
 

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