Question

I think that it is because the former is an array of pointers to char and the latter is a pointer to an array of chars, and we need to properly specify the size of the object being pointed to for our function definition. In the former;

function(char * p_array[])

the size of the object being pointed to is already included (its a pointer to char), but the latter

function(char (*p_array)[])

needs the size of the array p_array points to as part of p_array’s definition?
I’m at the stage where I’ve been thinking about this for too long and have just confused myself, someone please let me know if my reasoning is correct.

Answer 1

Both are valid in C but not C++. You would ordinarily be correct:

char *x[]; // array of pointers to char
char (*y)[]; // pointer to array of char

However, the arrays decay to pointers if they appear as function parameters. So they become:

char **x; // Changes to pointer to array of pointer to char
char (*y)[]; // No decay, since it's NOT an array, it's a pointer to an array

In an array type in C, one of the sizes is permitted to be unspecified. This must be the leftmost one (whoops, I said rightmost at first). So,

int valid_array[][5]; // Ok
int invalid_array[5][]; // Wrong

(You can chain them… but we seldom have reason to do so…)

int (*convoluted_array[][5])[][10];

There is a catch, and the catch is that an array type with [] in it is an incomplete type. You can pass around a pointer to an incomplete type but certain operations will not work, as they need a complete type. For example, this will not work:

void func(int (*x)[])
{
    x[2][5] = 900; // Error
}

This is an error because in order to find the address of x[2], the compiler needs to know how big x[0] and x[1] are. But x[0] and x[1] have type int [] — an incomplete type with no information about how big it is. This becomes clearer if you imagine what the “un-decayed” version of the type would be, which is int x[][] — obviously invalid C. If you want to pass a two-dimensional array around in C, you have a few options:

• Pass a one-dimensional array with a size parameter.

  void func(int n, int x[])
  {
      x[2*n + 5] = 900;
  }
  

• Use an array of pointers to rows. This is somewhat clunky if you have genuine 2D data.

  void func(int *x[])
  {
      x[2][5] = 900;
  }
  

• Use a fixed size.

  void func(int x[][5])
  {
      x[2][5] = 900;
  }
  

• Use a variable length array (C99 only, so it probably doesn’t work with Microsoft compilers).

  // There's some funny syntax if you want 'x' before 'width'
  void func(int n, int x[][n])
  {
      x[2][5] = 900;
  }
  

This is a frequent problem area even for C veterans. Many languages lack intrinsic “out-of-the-box” support for real, variable size, multidimensional arrays (C++, Java, Python) although a few languages do have it (Common Lisp, Haskell, Fortran). You’ll see a lot of code that uses arrays of arrays or that calculates array offsets manually.