Trying to assign a pointer value A with type "char * const *" into a variable B declared as "char const * const * " triggers a warning, although the const-yness of the type is _increased_ not decreased.

Such a thing may be needed when trying to call a function that displays a list of strings:

void displayList(char const * const * list)
{
   /* display list */
}

The char const * declaration is needed if somewhere in the program, this function is called with an array of static strings:

   char const  *list1[] = { "hello", "world", NULL };
   displayList(list1);


However, if elsewhere in the program, we want to call the function with a list of _dynamically_ allocated strings, we get a "initialization from incompatible pointer type" warning:
   char **list2;
   for(i=0; i< N; i++)
      list2[i] = strdup( some expression );

   ...
   displayList(list2);
  
   for(i=0; i<N; i++)
      free(list2[i]);

Because of the free(), we cannot declare list2[] as char const ** , but then this restriction on _adding_ the const qualifier during assignment or parameter passing is causing trouble.


There is a very similar case, where such a restriction on adding a const qualifier makes sense:

  assiging  char **  to  char const **

(i.e. the case where in the target, only the string _contents_ are const, but not the string pointers themselves).

In this case, the assignment warning would be appropriate, in order to prevent the following:

char *launderConst(char const *in)
{
    char * source[1];
    char const ** target = source;

    target[0] = in;
    return source[0];
}


However if the target list itself is const (rather than just its string contents), we would never be able to do this:
    char const *const* target = source;
    target[0] = in;

So why the warning about the first line, if the second one is forbidden anyways (assigning a value to a const)?


Or is there another way of declaring the displayList() function without triggering a warning in any of both use-cases?