Thread: Incompatible pointer types - C

Nothing you have in main at the moment can correctly call checkSizesLast() with any amount of casting or dereferencing.

checkSizesLast is expecting an array (of some dimensions) who's most basic elements are char pointers.

You can't make an array of pointers-to-T out of an array of T just by casting.

Code:

int main()
{
    char matrix[10][20][30];
    char matrix2[1][20][30];
    char *m3[20][30];
    checkSizes(matrix);
    checkSizesAgain(matrix);
    checkSizesLast(m3);
    checkSizes(matrix2);
    return 0;
}

Do you understand that these all say essentially the same thing?

Code:

void f(char   matrix3d [10][20][30]);
void g(char   matrix3d [  ][20][30]);
void h(char (*matrix3d)    [20][30]);

When you pass an array to a function, the outermost dimension "decays" into a pointer, losing the size information for that dimension.

@john.c
Yea yea I know that - thank you for the highlight - appreciate it.

BTW - is the third one equivalent to the first two statements?

EDIT:
You have said they are the same thing.
Umm... but to me it seems like the third one sends a pointer to an array of matrix 20X30 of chars.... It looks like there is no reason to decay anything here as matrix3d is already decayed in some manner

Yes, they are all the same. The point is that you can pass the exact same array to all of them. The last version is just explicitly saying what is actually happening. The other two are in a sense somewhat misleading, especially the first, which seems to suggest that the 10 has some meaning whereas it has no meaning at all. It is completely ignored.

Code:

#include <stdio.h>
 
void f(char   m [10][20][30]) { printf("%c\n", m[0][0][0]); }
void g(char   m [  ][20][30]) { printf("%c\n", m[0][0][0]); }
void h(char (*m)    [20][30]) { printf("%c\n", m[0][0][0]); }
 
int main() {
    char m[10][20][30];
    m[0][0][0] = 'x';
    f(m);
    g(m);
    h(m);   // no complaints !
    return 0;
}