jovanovic.dragoslav
BAN USER#include <stdio.h>
int main(){
int i, j;
int *p, *q;
int **x,
***y;
i = 100;
j = 200;
p = &i;
q = &j;
x = &p;
y = &x;
*p = *p + *q;
*q = **x / 2;
**x = *p + j;
printf(" i = %d\n", i);
printf("&i = %p\n", &i);
printf(" j = %d\n", j);
printf("&j = %p\n\n", &j);
printf(" p = %p\n", p);
printf("&p = %p\n", &p);
printf("*p = %d\n", *p);
printf(" q = %p\n", q);
printf("&q = %p\n", &q);
printf("*q = %d\n\n", *q);
printf(" x = %p\n", x);
printf("&x = %p\n", &x);
printf("*x = %p\n", *x);
printf("**x= %d\n\n", **x);
***y = ***y +1;
printf(" i = %d\n", i);
printf(" y = %p\n", y);
printf("&y = %p\n", &y);
printf("*y = %p\n", *y);
printf("**y= %p\n", **y);
printf("***y= %d\n", ***y);
return 0;
}
This should be the answer.
I took the liberty of modify the above example a bit. There is no need for an output buffer. We only have to remember that "+" is a string i.e. two bytes '+; and '\0' and 'x' is one byte, therefore we can simply replace the current buffer[i] with '+' in case of vowel. I kept the copyright notice.
- jovanovic.dragoslav June 30, 2016