Array Partition with pointer - c++

Question

I have a 20 x 20 array and need to iterate over it by reading a 4 x 4 array. I thought I could do this with pointed assignment, but it does not do much except force close
const char SOURCE[20][20];
const char **pointer;
for(int x = 0; x < 20; x+=4)
{
for(int y = 0; y < 20; y+=4)
{
pointer = (const char **)&SOURCE[x][y];
printGrid(pointer);
}
}
void printGrid(const char **grid)
{
// do something usefull
}

Answer 1

Just casting a pointer to a different type doesn't change the
type of what it points to (and will usually lead to undefined
behavior, unless you really know what you're doing). If you
cannot change printGrid, you'll have to create an array of
pointers on the fly:
for ( int x = 0; x < 20; x += 4 ) {
for ( int y = 0; y < 20; y += 4 ) {
char const* p4[4] =
{
source[x] + y,
source[x + 1] + y,
source[x + 2] + y,
source[x + 3] + y
};
printGrid( p4 );
}
}

Answer 2

A pointer to a pointer is not the same as an array of arrays.
You can however use a pointer to an array instead:
const char (*pointer)[20];
You of course need to update the printGrid function to match the type.
As for the reason why a pointer-to-pointer and an array-of-array (also often called a matrix) see e.g. this old answer of mine that shows the memory layout of the two.

Answer 3

Your 2D-array is of type char:
const char SOURCE[20][20];
When you are iterating through it, you can either look at the char or reference the address with a char*:
for(int x = 0; x < 20; x+=4)
{
for(int y = 0; y < 20; y+=4)
{
printGrid(SOURCE[x][y]); // do this unless you need to do something with pointer
}
}
Then you can make printGrid with either of the following signatures:
void printGrid(const char& grid)
{
// do something usefull
}
or
void printGrid(const char* grid)
{
// do something usefull
}

Answer 4

Extending James's Answer, you may change your code as below as it sees that the it passes pointer to an array of 4 char rather than just array of char.
for(int x = 0; x < 20; x+=4)
{
for(int y = 0; y < 20; y+=4)
{
char const (*p4[4])[4] =
{
(const char(*)[4])(SOURCE[x] + y),
(const char(*)[4])(SOURCE[x + 1] + y),
(const char(*)[4])(SOURCE[x + 2] + y),
(const char(*)[4])(SOURCE[x + 3] + y)
};
}
}