Hi, i have such situation:
but i get a warning:Code:#include <stdio.h> typedef int x[2][2]; int y(int **i) { i[0][1]=1; return 0; } int main() { x one; y(&one); printf("%d\n",one[0][1]); return 0; }
$ gcc hhhh.c
hhhh.c: In function `main':
hhhh.c:15: warning: passing arg 1 of `y' from incompatible pointer type
and of course, printed value is bad.
What's wrong with that?
Regards.