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    <p>This code will <em>not</em> work in either C or C++. An array of type <code>int[4][4]</code> is not convertible to a pointer of type <code>int **</code> (which is what <code>int *arr[]</code> stands for in parameter declaration). If you managed to <em>compile</em> it in C, it is simply because you probably ignored a C compiler warning of basically the same format as the error message you got from C++ compiler. (Sometimes C compilers issue <em>warnings</em> for what is essentially an <em>error</em>.)</p> <p>So, again, don't make assertions that are not true. This code does not work in C. In order to convert a built-in 2D array into a <code>int **</code> pointer you can use a technique like this one</p> <p><a href="https://stackoverflow.com/questions/1584100/converting-multidimensional-arrays-to-pointers-in-c">Converting multidimensional arrays to pointers in c++</a></p> <p>(See the accepted answer. The problem is exactly the same.)</p> <p><strong>EDIT:</strong> The code <em>appears</em> to work in C because another bug in the printing code is masquerading the effects of the bug in array passing. In order to properly access an element of an <code>int **</code> pseudo-array, you have to use expression <code>*(*(arr + i) + j)</code>, or better a plain <code>arr[i][j]</code> (which is the same thing). You missed the extra <code>*</code> which made it print something that has absolutely nothing to do with the content of your array. Again, initialize your array in <code>main</code> to something else to see that the results you are printing in C have absolutely nothing to do with the your intended content of the array.</p> <p>If you change the <code>printf</code> statement as shown above, your code will most likely crash because of the array-passing bug I described initially.</p> <p>One more time: you cannot pass a <code>int[4][4]</code> array as an <code>int **</code> pseudo-array. This is what the C++ is telling you in the error message. And, I'm sure, this is what your C compiler told you, but you probably ignored it, since it was "just a warning".</p>
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    1. POHow to pass a multidimensional array to a function in C and C++
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    1. COhttp://codepad.org/ZKBvaGiW
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    2. CO@cambr: One more time: the code does not work. Whatever results you are getting from it in C is just garbage, which just accidentally happens to look like the correct output. Initialize your array to something else (not zeros) and try again.
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    3. COWell. THe type is different, but the underlying memory address is the same. Hence it is rational to treat this as a warning for a C compiler.
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