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POWhy do I get different results when I dereference a pointer after freeing it?
 

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  1. POWhy do I get different results when I dereference a pointer after freeing it?
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    <p>I've a question about the memory management in C (and GCC 4.3.3 under Debian GNU/Linux). </p> <p>According to the C Programming Language Book by K&amp;R, (chap. 7.8.5), when I free a pointer and then dereference it, is an error. But I've some doubts since I've noted that sometimes, as in the source I've pasted below, the compiler (?) seems to work according a well-defined principle. </p> <p>I've a trivial program like this, that shows how to return an array dynamically allocated:</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;stdlib.h&gt; int * ret_array(int n) { int * arr = (int *) malloc(10 * sizeof(int)); int i; for (i = 0; i &lt; n; i++) { arr[i] = i*2; } printf("Address pointer in ret_array: %p\n", (void *) arr); return arr; } int * ret_oth_array(int n) { int * arr = (int *) malloc(10 * sizeof(int)); int i; for (i = 0; i &lt; n; i++) { arr[i] = i+n; } printf("Address pointer in ret_oth_array: %p\n", (void *) arr); return arr; } int main(void) { int *p = NULL; int *x = NULL; p = ret_array(5); x = ret_oth_array(6); printf("Address contained in p: %p\nValue of *p: %d\n", (void *) p, *p); free(x); free(p); printf("Memory freed.\n"); printf("*(p+4) = %d\n", *(p+4)); printf("*x = %d\n", *x); return 0; } </code></pre> <p>If I try to compile it with some arguments: <code>-ansi -Wall -pedantic-errors</code>, it doesn't raises errors or warning. Not only; it also runs fine.</p> <pre><code>Address pointer in ret_array: 0x8269008 Address pointer in ret_oth_array: 0x8269038 Address contained in p: 0x8269008 Value of *p: 0 Memory freed. *p+4 = 8 *x = 0 </code></pre> <p>*(p+4) is 8 and *x is 0. Why does this happen? If *(p+4) is 8, shouldn't *x be 6, since the first element of the x-array is 6?</p> <p>Another strange thing happens if I try to change the order of the calls to free. E.g.:</p> <pre><code>int main(int argc, char * argv[]) { /* ... code ... */ free(p); free(x); printf("Memory freed.\n"); printf("*(p+4) = %d\n", *(p+4)); printf("*x = %d\n", *x); return 0; } </code></pre> <p>In fact in this case the output (on my machine) will be:</p> <pre><code>*p+4 = 8 *x = 142106624 </code></pre> <p>Why does the <em>x</em> pointer is really "freed", while the p pointer is freed (I hope) "differently"? Ok, I know that after freeing memory I should make the pointers to point to NULL, but I was just curious :P</p>
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    1. COTry to run more than once and then compare the results
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      1. POWhy do I get different results when I dereference a pointer after freeing it?
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    2. COI've run it at least ten times and I always get the same results
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    3. COIn a single-threaded program on the same machine, these things often behave reproducibly - that's why they lead to "works-on-my-machine"-kind of bugs. However, add some code before/after it, add another thread, use different compiler settings and you will get different results
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