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    <p>Take it easy. It is only a compiler error. Arrays are pretty tricky. Here is the <em>rule</em>:</p> <blockquote> <p>The value of a variable of type array <strong>decays</strong> to the <strong>address of element zero</strong> of this array</p> </blockquote> <p>Your first snippet looks like:</p> <pre><code>int a[2] = {1,2}; </code></pre> <p>So according to the rule if <code>a</code> is in the right hand side of a assignment then it decays to address of the element zero and that is why it has type <code>int *</code>. This brings you to</p> <pre><code>int *b = a; </code></pre> <p>In the second snippet what you really have is an array of arrays. (By the way, to make it explicit I've changed your code a bit.)</p> <pre><code>int a[2][2]={{1,2},{3,4}}; </code></pre> <p>This time <code>a</code> will decay to the pointer to an <em>array of two integers</em>! So if you would want to assign <code>a</code> to something, you would need this something to have the same type.</p> <pre><code>int (*b)[2] = a; //Huh! </code></pre> <p>(This syntax maybe a bit stunning to you, but just think for a moment that we have written <code>int *b[2];</code> Got the point? <code>b</code> would be an array of pointers to integers! Not really what we wanted...)</p> <p>You could stop reading here, but you could also move on, because I have not told you all the truth. The <em>rule</em> I mentioned has three exceptions...</p> <p>The value of the array will <strong>not</strong> decay to the address of the element zero <strong>if</strong></p> <ol> <li>array is operand of <code>sizeof</code></li> <li>array is operand of <code>&amp;</code></li> <li>array is a literal string initializer for a character array</li> </ol> <p>Let's explain these exceptions in more detail and with examples:</p> <pre><code>int a[2]; int *pi = a ; /* the same as pi = &amp;a[0]; */ printf("%d\n", sizeof(a)); /* size of the array, not of a pointer is printed! */ int (*pi2)[2] = &amp;a; /* address of the array itself is taken (not the address of a pointer) */ </code></pre> <p>And finally</p> <pre><code>char a[] = "Hello world "; </code></pre> <p>Here not a pointer to "Hello world" is copied, but the whole string is copied and a points to this copy.</p> <p>There is <em>really</em> a lot of information and it is really difficult to understand everything at once, so take your time. I recommend you to read K&amp;R on this topic and afterwards <a href="http://rads.stackoverflow.com/amzn/click/0131774298">this</a> excellent book.</p>
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    1. POWhat is difference between types int** and int[][]?
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    1. USBeginner
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    1. CO+1, calming the developer when debugging in C++ is usually the first step. :D
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    2. COIt's not a warning, it's a fundamental incompatibility in the binary layout of the two types.
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      1. USChris Lutz
    3. CO@ChrisLutz Thank you, just my bad English.
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