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copied!<p>No, the behavior is not defined. Moreover, the code is not supposed to compile. </p> <p>Firstly, the code is not supposed to compile because it contains a constraint violation. The expression you are passing as an operand to <code>free</code> has <code>int</code> type. The parameter of <code>free</code> has <code>void *</code> type. The only case when an <code>int</code> value can be implicitly converted to <code>void *</code> type is when the <code>int</code> value is an Integral Constant Expression (ICE) with value <code>0</code>. In your case <code>x</code> is not an ICE, meaning that it is not implicitly convertible to <code>void *</code>. The only reason your code compiles is that for historical reasons (to support legacy code) your compiler quietly overlooks the constraint violation present in the <code>free(x)</code> call. I'm sure that if you elevate the level of warnings in your compiler, it will complain (at least with a warning). A pedantic compiler will immediately issue an error for <code>free(x)</code> call. Try Comeau Online, for example in C89/90 mode:</p> <pre><code>"ComeauTest.c", line 6: error: argument of type "int" is incompatible with parameter of type "void *" free(x); ^ </code></pre> <p>(Also, did you remember to include <code>stdlib.h</code> before calling <code>free</code>?)</p> <p>Secondly, let's assume that the code compiles, i.e. it is interpreted by the compiler as <code>free((void *) x)</code>. In this case a non-constant integral value <code>x</code> is converted to pointer type <code>void *</code>. The result of this conversion is implementation defined. Note, that the language guarantees that when an ICE with value of <code>0</code> is converted to pointer type, the result is a null pointer. But in your case <code>x</code> is not an ICE, so the result of the conversion is implementation-defined. In C there's no guarantee that you will obtain a null pointer by converting a non-ICE integer with value <code>0</code> to pointer type. On your implementation it probably just happened that <code>(void *) x</code> with non-ICE <code>x</code> equal to <code>0</code> produces a null pointer value of type <code>void *</code>. This null pointer value, when passed to <code>free</code>, results in a no-op, per the specification of <code>free</code>. </p> <p>In general case though, passing such a pointer to <code>free</code> will result in undefined behavior. The pointers that you can legally pass to <code>free</code> are pointers obtained by previous calls to <code>malloc</code>/<code>calloc</code>/<code>realloc</code> and null pointers. Your pointer violates this constraint in general case, so the behavior is undefined.</p> <p>This is what happens in your case. But, again, your code contains a constraint violation. And even if you override the violation, the behavior is undefined.</p> <p><strong>P.S.</strong> Note, BTW, that many answers already posted here make the same serious mistake. They assume that <code>(void *) x</code> with zero <code>x</code> is supposed to produce a null pointer. This is absolutely incorrect. Again, the language makes absolutely no guarantees about the result of <code>(void *) x</code> when <code>x</code> is not an ICE. <code>(void *) 0</code> is guaranteed to be null pointer, but <code>(void *) x</code> with zero <code>x</code> is <em>not</em> guaranteed to be null pointer.</p> <p><strong>This is covered in C FAQ</strong> <a href="http://c-faq.com/null/runtime0.html" rel="nofollow noreferrer">http://c-faq.com/null/runtime0.html</a> . For those interested in better understanding of why it is so it might be a good idea to read the entire section on null pointers <a href="http://c-faq.com/null/index.html" rel="nofollow noreferrer">http://c-faq.com/null/index.html</a></p>

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