const array const {}

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 c开发者_开发技巧an 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?

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.

Use cdecl. It gives an error on the second entry. The first only clearly suggests that the second const refers to the *.

In C/C++, you cannot pass an entire array as an argument to a function. You can, however, pass to the function a pointer to an array by specifying the array's name without an index.

(E.g) This program fragment passes the address of i to func1() :

int main(void)
{
int i[10];
func1(i);
.
.
.
}

To receive i, a function called func1() can be defined as

void func1(int x[]) /* unsized array */
{
.
.
}

or

void func1(int *x) /* pointer */
{
.
.
}

or

void func1(int x[10]) /* sized array */
{
.
.
}

source : THE COMPLETE REFERENCE - HERBERT.