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2528999
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2010-03-27T10:57:42.970
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2010-03-27T10:57:42.970
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In C, when you used the name of an array in an expression (including passing it to a function), unless it is the operand of the address-of (<code>&</code>) operator or the <code>sizeof</code> operator, it decays to a pointer to its first element. That is, in most contexts <code>array</code> is equivalent to <code>&array[0]</code> in both type and value. In your example, <code>my_array</code> has type <code>char[100]</code> which decays to a <code>char*</code> when you pass it to printf. <code>&my_array</code> has type <code>char (*)[100]</code> (pointer to array of 100 <code>char</code>). As it is the operand to <code>&</code>, this is one of the cases that <code>my_array</code> doesn't immediately decay to a pointer to its first element. The pointer to the array has the same address value as a pointer to the first element of the array as an array object is just a contiguous sequence of its elements, but a pointer to an array has a different type to a pointer to an element of that array. This is important when you do pointer arithmetic on the two types of pointer. <code>pointer_to_array</code> has type <code>char *</code> - initialized to point at the first element of the array as that is what <code>my_array</code> decays to in the initializer expression - and <code>&pointer_to_array</code> has type <code>char **</code> (pointer to a pointer to a <code>char</code>). Of these: <code>my_array</code> (after decay to <code>char*</code>), <code>&my_array</code> and <code>pointer_to_array</code> all point directly at either the array or the first element of the array and so have the same address value.
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1. POHow come an array's address is equal to its value in C?
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