an advanced declaration of a function? [duplicate] - c++

Question

For reading complex pointer declarations there is the right-left rule.
But this rule does not mention how to read const modifiers.
For example in a simple pointer declaration, const can be applied in several ways:
char *buffer; // non-const pointer to non-const memory
const char *buffer; // non-const pointer to const memory
char const *buffer; // equivalent to previous declartion
char * const buffer = {0}; // const pointer to non-const memory
char * buffer const = {0}; // error
const char * const buffer = {0}; // const pointer to const memory
Now what about the use of const with a pointer of pointer declaration?
char **x; // no const;
const char **x;
char * const *x;
char * * const x;
const char * const * x;
const char * * const x;
const char * const * const x;
And what is an easy rule to read those declarations?
Which declarations make sense?
Is the Clockwise/Spiral Rule applicable?
Two real world examples
The method ASTUnit::LoadFromCommandLine uses const char ** to supply command line arguments (in the llvm clang source).
The argument vector parameter of getopt() is declared like this:
int getopt(int argc, char * const argv[], const char *optstring);
Where char * const argv[] is equivalent to char * const * argv in that context.
Since both functions use the same concept (a vector of pointers to strings to supply the arguments) and the declarations differ - the obvious questions are: Why do they differ? Makes one more sense than the other?
The intend should be: The const modifier should specify that the function does not manipulate strings of this vector and does not change the structure of the vector.

Answer 1

The const modifier is trivial: it modifies what precedes it, unless
nothing precedes it. So:
char const* buffer; // const modifies char
char* const buffer; // const modifies *
, etc. Generally, It's best to avoid the forms where nothing precedes
the const, but in practice, you're going to see them, so you have to
remember that when no type precedes the const, you have to logically
move it behind the first type. So:
const char** buffer;
is in fact:
char const** buffer;
, i.e. pointer to pointer to const char.
Finally, in a function declaration, a [] after reads as a * before.
(Again, it's probably better to avoid this misleading notation, but
you're going to see it, so you have to deal with it.) So:
char * const argv[], // As function argument
is:
char *const * argv,
a pointer to a const pointer to a char.

Answer 2

(Trying to focus on other aspects of the question)
The rule of thumb for const declarations is to read them from right to left and const modifies the next token. Exception: At the beginning of a declaration const modifies the previous token.
There is a rationale behind this exception - for elementary declarations const char c looks for some people more natural than char const c - and it is reported that a precursor form of const char c predates the final const rule.
getopt
int getopt(int argc, char * const argv[], const char *optstring);
or
int getopt(int argc, char * const * argv, const char *optstring);
Which means that argv is a pointer to const vector of pointers to non-const strings.
But one would expect following declaration:
int getopt(int argc, char const * const * argv, const char *optstring);
(pointer to const vector to const strings)
Because getopt() is not supposed to change the strings referenced via argv.
At least char ** (as used in main()) automatically converts to char * const * argv.
Clang
ASTUnit::LoadFromCommandLine(..., const char **argv, ...);
Which means that argv is a pointer to a non-const array of pointers to const strings.
Again one would expect const char * const *argv for the same reason as above.
But this is more noticeable because char ** does not convert to const char **, e.g.
int main(int argc, char **argv) {
const char **x = argv; // Compile error!
return 0;
}
yields a compile error, where
int main(int argc, char **argv) {
char * const *x = argv;
return 0;
}
and
int main(int argc, char **argv) {
const char * const *x = argv;
return 0;
}
do not.