Question

So you can do this:

void foo(const int * const pIntArray, const unsigned int size);

Which says that the pointer coming is read-only and the integer’s it is pointing to are read-only.

You can access this inside the function like so:

blah = pIntArray[0]

You can also do the following declaration:

void foo(const int intArray[], const unsigned int size);

It is pretty much the same but you could do this:

intArray = &intArray[1];

Can I write:

void foo(const int const intArray[], const unsigned int size);

Is that correct?

Answer 1

No, your last variant is not correct. What you are trying to do is achieved in C99 by the following new syntax

void foo(const int intArray[const], const unsigned int size);

which is equivalent to

void foo(const int *const intArray, const unsigned int size);

That [const] syntax is specific to C99. It is not valid in C89/90.

Keep in mind that some people consider top-level cv-qualifiers on function parameters “useless”, since they qualify a copy of the actual argument. I don’t consider them useless at all, but personally I don’t encounter too many reasons to use them in real life.