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2013-12-11T02:42:28.430
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<pre><code>char *p = "String"; </code></pre> This is legal, but it really should be: <pre><code>const char *p = "String"; </code></pre> so you don't accidentally try to modify the array. What array, you ask? Any string literal corresponds to a statically allocated array of size <code>LEN+1</code>, where <code>LEN</code> is the length of the literal. So in this case, there's an anonymous array of type <code>char[7]</code> that exists during the entire execution of your program, containing the values <pre><code>`{ 'S', 't', 'r', 'i', 'n', 'g', '\0' }`. </code></pre> The initializer for <code>p</code> causes it to point to the first element of that array, which contains the <code>'S'</code>. And since the array was not allocated by a call to <code>malloc()</code> (or <code>calloc()</code>, or <code>realloc()</code>), you must not attempt to <code>free()</code> it. Given <code>p</code>, you can't directly determine the size of the array (since <code>p</code> points to its first element, not to the array as a whole). You can compute <code>strlen(p) + 1</code>, which is probably the same as the allocated size -- but if the string literal were <code>"foo\0bar"</code>, then it would be 8 bytes long, but <code>strlen</code> would stop at the first null character and return <code>3</code>. So if you need to be able to determine the size of the array, don't discard that information by assigning its address to a pointer. Now if you had instead written: <pre><code>char arr[] = "String"; </code></pre> the initializer would copy the contents of the string literal into the array <code>arr</code>, and the size of <code>arr</code> would be determined by the compiler from the size of the initializer. Again, you must not attempt to call <code>free()</code> on this array (or rather, on a pointer to its first element), since it wasn't allocated with <code>malloc()</code>. But you can easily determine its size via <code>sizeof arr</code>, which will yield <code>7</code>. (<code>sizeof p</code> would give you the size of a pointer, not of the array it points to.) Finally, if you used <code>malloc()</code> to allocate an array: <pre><code>#define message "String" char *p = malloc(sizeof message + 1); if (p == NULL) { fprintf(stderr, "malloc failed\n"); exit(EXIT_FAILURE); } strcpy(p, message); </code></pre> you still can't use <code>p</code> (which is just a pointer to a single character) to determine the size of the allocated array -- but <code>free()</code> is able to figure it out. How? The language doesn't say, but any implementation will keep extra bookkeeping information behind the scenes to let <code>free()</code> do the right thing. Note that the implementation doesn't necessarily remember the size you requested. A call to <code>malloc(7)</code> might actually allocate, say, 16 bytes, and the only information <code>free</code> needs is the base address and the actual allocated size. But since you wrote the call to <code>malloc</code>, if you need to remember the allocated size, just save it somewhere: <pre><code>// ... const size_t allocated_bytes = sizeof message + 1; char *p = malloc(allocated_bytes); // ... </code></pre> If you have a declared array object, you can use <code>sizeof</code> to determine its size. Arrays are usually manipulated using pointers, which do not retain the size of the array; you have to keep track of that yourself. <code>malloc()</code> and friends do maintain some size information behind the scenes, but only for the purpose of letting <code>free()</code> work correctly; you (probably) can't access that information yourself. If you need to keep track of the size of an allocated array, you have to do it yourself.
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1. POHow does size information get passed with a pointer?
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1. USKeith Thompson
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