I have two 2d arrays (pointer to pointer to unsigned) and I want to swap them. First I started to write the code for 1d array of pointers. This works perfectly:
#include <stdio.h>
#include <stdlib.h>
void swap(unsigned **a, unsigned **b) {
unsigned * tmp = *a;
*a = *b;
*b = tmp;
}
int main() {
size_t x;
unsigned *a = (unsigned*) malloc(10*sizeof(unsigned));
unsigned *b = (unsigned*) malloc(10*sizeof(unsigned));
for(x=0;x<10;x++) a[x] = 1;
for(x=0;x<10;x++) b[x] = 0;
printf("%u %u\n",a[5],b[5]);
swap(&a, &b);
printf("%u %u\n",a[5],b[5]);
return 0;
}
I thought I can do something similar for 2d arrays. Here is my try:
#include <stdio.h>
#include <stdlib.h>
void swap(unsigned ***a, unsigned ***b) {
unsigned ** tmp = **a;
**a = **b;
**b = tmp;
}
int main() {
size_t x,y;
unsigned **a = (unsigned**) malloc(10*sizeof(unsigned*));
unsigned **b = (unsigned**) malloc(10*sizeof(unsigned*));
for(x=0;x<10;x++)
{
a[x] = malloc(10*sizeof(unsigned));
b[x] = malloc(10*sizeof(unsigned));
}
for(x=0;x<10;x++) for(y=0;y<10;y++) a[x][y] = 1;
for(x=0;x<10;x++) for(y=0;y<10;y++) b[x][y] = 0;
printf("%u %u\n",a[5][5],b[5][5]);
swap(&a, &b);
printf("%u %u\n",a[5][5],b[5][5]);
return 0;
}
I got two compiler warnings:
$ gcc -g -Wall test.c
test.c: In function ‘swap’:
test.c:5:21: warning: initialization from incompatible pointer type [enabled by default]
test.c:7:7: warning: assignment from incompatible pointer type [enabled by default]
I tried to understand the warnings, but I still do not understand them. I have no idea what is wrong in my code.
unsigned ** tmp = *a; *a = *b; *b = tmp;