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    <p>&gt; However, in pure C, this still gives a warning, and I don't understand why</p> <p>You've already identified the problem -- this code is not const-correct. "Const correct" means that, except for const_cast and C-style casts removing const, you can never modify a const object through those const pointers or references.</p> <p>The value of const-correctness -- const is there, in large part, to detect programmer errors. If you declare something as const, you're stating that you don't think it should be modified -- or at least, those with access to the const version only should not be able to modifying it. Consider:</p> <pre><code>void foo(const int*); </code></pre> <p>As declared, foo doesn't have <em>permission</em> to modify the integer pointed to by its argument. </p> <p>If you're not sure why the code you posted isn't const-correct, consider the following code, only slightly different from HappyDude's code:</p> <pre><code>char *y; char **a = &amp;y; // a points to y const char **b = a; // now b also points to y // const protection has been violated, because: const char x = 42; // x must never be modified *b = &amp;x; // the type of *b is const char *, so set it // with &amp;x which is const char* .. // .. so y is set to &amp;x... oops; *y = 43; // y == &amp;x... so attempting to modify const // variable. oops! undefined behavior! cout &lt;&lt; x &lt;&lt; endl; </code></pre> <p>Non-const types can only convert to const types in particular ways to prevent any circumvention of 'const' on a data-type without an explicit cast. </p> <p>Objects initially declared const are particularly special -- the compiler can assume they never change. However, if 'b' can be assigned the value of 'a' without a cast, then you could inadvertantly attempt to modify a const variable. This would not only break the check you asked the compiler to make, to disallow you from changing that variables value -- it would also allow you break the compiler optimizations!</p> <p>On some compilers, this will print '42', on some '43', and others, the program will crash. </p> <p><strong>Edit-add:</strong></p> <p>HappyDude: Your comment is spot on. Either the C langauge, or the C compiler you're using, treats const char * const * fundamentally differently than the C++ language treats it. Perhaps consider silencing the compiler warning for this source line only.</p> <p><strong>Edit-delete:</strong> removed typo</p>
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    1. POWhy can't I convert 'char**' to a 'const char* const*' in C?
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    1. COWhat you have said is correct, but it does not address the question. As I said, I understand why converting from a char** to a const char** is incorrect. What I don't understand is why converting from a char** to a const char* const * is incorrect. I have updated my question to elucidate.
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    2. COIt's interesting that this response keeps getting voted up. I guess a lot of people don't read past the first couple lines of the question, something I'll keep in mind for the future. Though this is quite a well written post on a relatively obscure point, I just wish it was relevant :P.
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    3. CO@Aaron: your post isn't particularly relevant to the topic, but well written nonetheless.
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