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POC char array as pointers
 

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  1. POC char array as pointers
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    <p>I want to understand:</p> <ul> <li>why it happens that sometimes a <code>char[1]</code> in C is used as <code>char*</code> (why doing this?) and</li> <li>how the internals works (what's going on)</li> </ul> <p>Giving following sample program:</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;string.h&gt; struct test_struct { char *a; char b[1]; } __attribute__((packed)); ; int main() { char *testp; struct test_struct test_s; testp = NULL; memset(&amp;test_s, 0, sizeof(struct test_struct)); printf("sizeof(test_struct) is: %lx\n", sizeof(struct test_struct)); printf("testp at: %p\n", &amp;testp); printf("testp is: %p\n", testp); printf("test_s.a at: %p\n", &amp;test_s.a); printf("test_s.a is: %p\n", test_s.a); printf("test_s.b at: %p\n", &amp;test_s.b); printf("test_s.b is: %p\n", test_s.b); printf("sizeof(test_s.b): %lx \n", sizeof(test_s.b)); printf("real sizeof(test_s.b): %lx \n", ((void *)(&amp;test_s.b) - (void *)(&amp;test_s.a)) ); return 0; } </code></pre> <p>I get the following output (OS X, 64bit):</p> <pre><code>sizeof(test_struct) is: 9 testp at: 0x7fff62211a98 testp is: 0x0 test_s.a at: 0x7fff62211a88 test_s.a is: 0x0 test_s.b at: 0x7fff62211a90 test_s.b is: 0x7fff62211a90 sizeof(test_s.b): 1 real sizeof(test_s.b): 8 </code></pre> <p>Looking at the memory addresses, one can see that even the struct is 9 bytes large, 16 bytes were allocated which seems to be caused by <code>char b[1]</code>. But I'm not sure if those extra bytes were allocated due to optimization/mem alignment reasons, or if this has to do with C's internal treatment of char arrays.</p> <p>A real world example can be seen in <code>&lt;fts.h&gt;</code>:</p> <p>`man 3 fts` shows the struct member `fts_name` as:</p> <pre><code> char *fts_name; /* file name */ </code></pre> <p>while /usr/include/fts.h defines the member as:</p> <pre><code> char fts_name[1]; /* file name */ </code></pre> <p>In the end, <code>fts_name</code> can really be used as a pointer to a C-string. For example, printing to stdout with <code>printf("%s", ent-&gt;fts_name)</code> works.</p> <p>So if a <code>char[1]</code> is really one byte large, it couldn't be used as a memory pointer on my 64bit machine. On the other hand, treating this as a full blown <code>char *</code> doesn't work either, as can be seen with the <code>test_s.b is</code> output above, which should show a NULL pointer then...</p>
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    1. COMain idea: I'm gonna disappoint you, but pointers are not arrays.
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    2. CO`real sizeof(test_s.b): 8` is **wrong**. Because `sizeof(char*)` is `8` and `sizeof(char)` is `1`. Thats why your structure is 9 byte.
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      1. USShiplu Mokaddim
    3. CORead section 6 of the [comp.lang.c FAQ](http://www.c-faq.com/); it's an *excellent* explanation of the (often confusing) relationship between arrays and pointers in C.
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      1. USKeith Thompson
 

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