Tom Lachecki

You may have used code such as [1]:

int<em> a = f(); const int</em> b = a; g(b);</p> <p>// Success

Here we obtain a pointer b to the same object that a points to, but we say that it is now immutable (through b, at least).

However, C++ does not allow a programmer to add constness more than one layer of indirection away [2]:

int<strong> a = f(); const int</strong> b = a; g(b);</p> <p>// error: invalid conversion from 'int<strong>' to 'const int</strong>'

Of course, you can still make the pointer itself const [3]:

int<strong> a = f(); int</strong> const b = a; g(b);</p> <p>// Success

So it turns out that you can add constness to the pointer [3], or to the immediate pointee [1], but you can't add constness to the pointee of the pointee [2].

The question is: why not? How does adding constness violate const-correctness? Answers on a postcard, please.

Update: As Kniht correctly pointed out in the comments (and, in fact, as the standard itself explains in a note), this is invalid for very good reason:

</p> <h1>include <cassert></h1> <p>int main() { char* p = 0;</p> <p>//char const<strong> a = &p; // not allowed, but let's pretend it is char const</strong> a = (char const**)&p; // instead force the cast to compile</p> <p>char const<em> orig = "original"; </em>a = orig; // type of <em>a is char const</em>, which is the type of orig, this is allowed</p> <p>assert(p == orig); // oops! char<em> points to a char const</em> }