Const correctness for array pointers?

Question

Someone made an argument saying that in modern C, we should always pass arrays to functions through an array pointer, since array pointers have strong typing. Example:

Answer 1

OP describes a function func() that has the following signature.

void func(size_t n, const int (*arr)[n])

OP wants to call it passing various arrays

#define SZ(a) (sizeof(a)/sizeof(a[0]))

int array1[3];
func(SZ(array1), &array1);  // problem

const int array2[3] = {1, 2, 3};
func(SZ(array2), &array2);

How do I apply const correctness to array pointers passed as parameters?

With C11, use _Generic to do the casting as needed. The cast only occurs when the input is of the acceptable non-const type, thus maintaining type safety. This is "how" to do it. OP may consider it "bloated" as it is akin to this. This approach simplifies the macro/function call to only 1 parameter.

void func(size_t n, const int (*arr)[n]) {
  printf("sz:%zu (*arr)[0]:%d\n", n, (*arr)[0]);
}

#define funcCC(x) func(sizeof(*x)/sizeof((*x)[0]), \
  _Generic(x, \
  const int(*)[sizeof(*x)/sizeof((*x)[0])] : x, \
        int(*)[sizeof(*x)/sizeof((*x)[0])] : (const int(*)[sizeof(*x)/sizeof((*x)[0])])x \
  ))

int main(void) {
  #define SZ(a) (sizeof(a)/sizeof(a[0]))

  int array1[3];
  array1[0] = 42;
  // func(SZ(array1), &array1);

  const int array2[4] = {1, 2, 3, 4};
  func(SZ(array2), &array2);

  // Notice only 1 parameter to the macro/function call
  funcCC(&array1);  
  funcCC(&array2);

  return 0;
}

Output

sz:4 (*arr)[0]:1
sz:3 (*arr)[0]:42
sz:4 (*arr)[0]:1

Alternatively code could use

#define funcCC2(x) func(sizeof(x)/sizeof((x)[0]), \
    _Generic(&x, \
    const int(*)[sizeof(x)/sizeof((x)[0])] : &x, \
          int(*)[sizeof(x)/sizeof((x)[0])] : (const int(*)[sizeof(x)/sizeof((x)[0])])&x \
    ))

funcCC2(array1);
funcCC2(array2);