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

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.

To be considered compatible, the source pointer should be const in the immediately anterior indirection level. So, this will give you the warning in GCC:

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

But this won't:

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

Alternatively, you can cast it:

char **a;
const char* const* b = (const char **)a;

You would need the same cast to invoke the function f() as you mentioned. As far as I know, there's no way to make an implicit conversion in this case (except in C++).

However, in pure C, this still gives a warning, and I don't understand why

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.

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:

void foo(const int*);

As declared, foo doesn't have permission to modify the integer pointed to by its argument.

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:

char *y;

char **a = &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 = &x; // the type of *b is const char *, so set it 
         //     with &x which is const char* ..
         //     ..  so y is set to &x... oops;
*y = 43; // y == &x... so attempting to modify const 
         //     variable.  oops!  undefined behavior!
cout << x << endl;

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.

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 inadvertently 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!

On some compilers, this will print 42, on some 43, and others, the program will crash.

Edit-add:

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.