I'm trying to print the array index using pointers arithmetic. Does anyone have any idea how to do it? in particular "j" I would like you to do it.
#include <stdio.h>
int main(void) {
int b[10] = {2, 8, 4, 7, 1, -45, 120, 78, 90, -6};
int *pb, j = 0;
for(pb = &b[0]; pb < &b[10];) {
printf("[%d] = %d\n", j, *pb++);
}
return 0;
}
Like this:
int main(void) {
int b[10] = { 2, 8, 4, 7, 1, -45, 120, 78, 90, -6 };
int *pb, j = 0;
for (pb = &b[0]; pb < &b[10]; pb++) {
printf("[%td] = %d\n", pb-b, *pb);
}
return 0;
}
In pointer arithmetic you can get the index difference with subtraction: pb-b is the index of the element that b points to in array pb. I also moved the *pb++ to the for loop increment in order to avoid an off-by-one-error.
Here is another option:
int b[10] = {2, 8, 4, 7, 1, -45, 120, 78, 90, -6};
for (int* pb = b; pb != b + sizeof(b) / sizeof(*b); pb++)
printf("[%d] = %d\n", pb - b, *pb);
pb - b yields a 64-bit value, which should be printed via %ld. Since we know that pb - b is at most 10, you can cast it to int and keep the %d%, o in other words: printf("[%d] = %d\n", (int)(pb - b), *pb);.long int. In any case, you can check pointer size of your platform via sizeof(void*).
j++in the loop