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
}
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 );
}
}
char const (*p4[4])[4]??
char const (*p4[4))[4] would be an array of pointer to an array; I couldn't initialize it without an explicit cast, because I don't have any char const[4] whose address I could initialize it with. And it wouldn't have the right type to pass to printGrid.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.
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
}
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)
};
}
}
printGrid expects a pointer to an array of 4 char const*, each of which points to 4 char. You're giving it a pointer to a single char const*, which points to an array of 20 - y char. Not at all the same thing.printGrid needs a pointer to an array[4] of pointer, so at some point, you need a char const *tmp[4]; (which I don't see in your code). Nothing else will do. And you pass it the address of the first element in that array (which would almost certainly be obtained by the implicit array to pointer conversion).printGrid (since p4 converts implicitly to a char const (**)[4], and not a char const**). The fact that you need the explicit conversions should be a sign that something isn't right.