Cannot cast array to pointer

Question

I have the following source:

#include 
using namespace std;

void main(int j)
{
    char arr[10][10];
    char** ptr;
    ptr = arr;
}


        

Answer 1

The existing answers, though correct, don't make it very clear that there is a fundamental reason (apart from the language rules) why you cannot cast char [10][10] to char **. Even if you force the cast by saying something like

char arr[2][2];
char ** ptr = (char **)arr;

it won't actually work.

The reason is that in C and C++ a two-dimensional array is laid out in memory as an array of arrays. That is, in C a two-dimensional array is laid out in memory as a single allocation,

arr -> arr[0][0]
       arr[0][1]
       arr[1][0]
       arr[1][1]

You'll notice that arr doesn't point to a char * but to arr[0][0] which is a char; therefore, while arr can be cast to a char *, it cannot be cast to a char **.

The correct forced cast would be

char arr[2][2];
char * ptr = (char *)arr;

If you don't want to force the cast (always a good idea if possible!) you would say

char arr[2][2];
char * ptr = arr[0];

or, to make the outcome clearer,

char arr[2][2];
char * ptr = &arr[0][0];

And you now have (in effect) a pointer to an array of 4 characters. [Proviso: I can't find anything in the C standard that prohibits an implementation from adding padding between two rows of an array, but I don't believe that any real-world implementations do so, and common coding practice depends on the assumption that there will be no such padding.]

If you really needed to convert arr to a char ** you would have to explicitly create an array of pointers:

char arr[2][2]
char * arr_ptrs[2];
char ** ptr;
arr_ptrs[0] = arr[0];
arr_ptrs[1] = arr[1];
ptr = arr_ptrs;

The C language could in principle do this for you automatically if you tried to cast a two-dimensional array to a pointer-to-pointer but that would violate the programmer's expectation that a cast does not have side-effects such as allocating memory.

In Java, by way of comparison, a two-dimensional array is always an array of pointers to arrays, so that the array

char arr[][] = { {'a', 'b'}, {'c', 'd'} };

is laid out in memory as three separate allocations, in arbitrary order and not necessarily adjacent,

arr -> arr[0]
       arr[1]

arr[0] -> arr[0][0]
          arr[0][1]

arr[1] -> arr[1][0]
          arr[1][1]

You will immediately notice that this requires more memory than the equivalent C array, and is slower to evaluate at runtime. On the other hand, it does allow the rows of an array to have different lengths.

Answer 2

Arrays are not pointers (I notice a lot of books tend to make you think this, though). They are something completely different. A pointer is a memory address while an array is a contiguous set of some data.

In some cases, an array can decay to a pointer to its first element. You can then use pointer arithmetic to iterate through the contiguous memory. An example of this case would be when passing an array to a function as a parameter.

What you probably want to do here is something like:

char arr[10];
char * i = &arr[0];

Obviously you'll need to use 2D arrays and char** in your case. I'll leave that to you to figure out :)