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    <p>The type of <code>a</code> is <code>int[2]</code>, not <code>int*</code>. Try again with</p> <pre><code>extern int a[2]; </code></pre> <hr> <p>The C compiler <strong>cannot type-check across source files</strong>. Therefore, when you say <code>int* a</code>, the compiler will assume you're telling the truth, use pointer semantics, and will not issue any compiler error.</p> <p>There are subtle difference between arrays and pointers. Let's assume a 32-bit system. Then the content of "a" will be distributed like this:</p> <pre> a 0x100 0x104 0x108 ← address +-----------+----------+ | 3 | 2 | ← content +-----------+----------+ </pre> <p>When <code>a</code> is an <strong>array</strong>,</p> <ul> <li>The value of the expression <code>a</code> is will be converted to the address of <code>a</code>. Therefore, when you print <code>a</code>, you will get its address, i.e. "0x100".</li> <li>The operation <code>a[n]</code> in C is equivalent to <code>*(a + n)</code>, i.e. advance the address <code>a</code> by <code>n</code> units, and then dereference it to get the content. Therefore, <code>a[0]</code> is equivalent to <code>*0x100</code>, which returns the content at 0x100, i.e. "3".</li> </ul> <p>When <code>a</code> is a <strong>pointer</strong>,</p> <ul> <li>The "value" of <code>a</code> is the content at the provided address. In fact this is the norm, the array type is a special case. Therefore, when you print <code>a</code>, you will get the content at that address, i.e. "3".</li> <li>The operation <code>a[n]</code> is still <code>*(a + n)</code>. Therefore, <code>a[0]</code> is equivalent to <code>*3</code>, which causes segmentation fault because the address "3" is invalid.</li> </ul>
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    1. POWhat is wrong with this C program using extern keyword?
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    1. POWhat is wrong with this C program using extern keyword?
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    1. COBut `int a[2]` is pointer, too, except the compiler just knows it's size at compile-time. Or is my understanding wrong here?
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    2. COBut this program can compile successfully, no error.
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    3. CO@NiklasR: No `int a[2]` is not a pointer, it's an array. `extern int a[];` would work as well so the difference is not the known size.
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