pointers - Why can't I convert 'char**' to a 'const char* const*' in C?

Question

The following code snippet (correctly) gives a warning in C and an error in C++ (using gcc & g++ respectively, tested with versions 3.4.5 and 4.2.1; MSVC does not seem to care):

char **a;
const char** b = a;

I can understand and accept this.
The C++ solution to this problem is to change b to be a const char * const *, which disallows reassignment of the pointers and prevents you from circumventing const-correctness (C++ FAQ).

char **a;
const char* const* b = a;

However, in pure C, the corrected version (using const char * const *) still gives a warning, and I don't understand why. Is there a way to get around this without using a cast?

To clarify:
1) Why does this generate a warning in C? It should be entirely const-safe, and the C++ compiler seems to recognize it as such.
2) What is the correct way to go about accepting this char** as a parameter while saying (and having the compiler enforce) that I will not be modifying the characters it points to? For example, if I wanted to write a function:

void f(const char* const* in) {
  // Only reads the data from in, does not write to it
}

And I wanted to invoke it on a char**, what would be the correct type for the parameter?

Answer 1

I had this same problem a few years ago and it irked me to no end.

The rules in C are more simply stated (i.e. they don't list exceptions like converting char** to const char*const*). Consequenlty, it's just not allowed. With the C++ standard, they included more rules to allow cases like this.

In the end, it's just a problem in the C standard. I hope the next standard (or technical report) will address this.