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copied!<p>(Trying to focus on other aspects of the question)</p> <p>The <a href="http://en.wikipedia.org/wiki/Const-correctness#C.2B.2B_syntax" rel="nofollow">rule of thumb for const declarations</a> is to read them from right to left and <code>const</code> modifies the next token. Exception: At the beginning of a declaration <code>const</code> modifies the previous token.</p> <p>There is a <a href="http://carlo17.home.xs4all.nl/cpp/const.qualifier.html" rel="nofollow">rationale behind this exception</a> - for elementary declarations <code>const char c</code> looks for some people more natural than <code>char const c</code> - and it is reported that a precursor form of <code>const char c</code> predates the final const rule.</p> <h2>getopt</h2> <pre><code>int getopt(int argc, char * const argv[], const char *optstring); </code></pre> <p>or</p> <pre><code>int getopt(int argc, char * const * argv, const char *optstring); </code></pre> <p>Which means that <code>argv</code> is a pointer to const vector of pointers to non-const strings.</p> <p>But one would expect following declaration:</p> <pre><code>int getopt(int argc, char const * const * argv, const char *optstring); </code></pre> <p>(pointer to const vector to const strings)</p> <p>Because <code>getopt()</code> is not supposed to change the strings referenced via argv.</p> <p>At least <code>char **</code> (as used in <code>main()</code>) automatically converts to <code>char * const * argv</code>.</p> <h2>Clang</h2> <pre><code>ASTUnit::LoadFromCommandLine(..., const char **argv, ...); </code></pre> <p>Which means that <code>argv</code> is a pointer to a non-const array of pointers to const strings.</p> <p>Again one would expect <code>const char * const *argv</code> for the same reason as above.</p> <p>But this is more noticeable because <code>char **</code> <a href="http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.17" rel="nofollow">does not convert</a> to <code>const char **</code>, e.g.</p> <pre><code>int main(int argc, char **argv) { const char **x = argv; // Compile error! return 0; } </code></pre> <p>yields a compile error, where</p> <pre><code>int main(int argc, char **argv) { char * const *x = argv; return 0; } </code></pre> <p>and</p> <pre><code>int main(int argc, char **argv) { const char * const *x = argv; return 0; } </code></pre> <p>do not.</p>

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