What does char*ptr=(char*)&i;exactly do in the following code
#include <stdio.h>
 int main()
 {
  int i=32;
  char*ptr=(char*)&i;
  printf("%d",*ptr);
  return 0;
 }
What does char*ptr=(char*)&i;exactly do in the following code
#include <stdio.h>
 int main()
 {
  int i=32;
  char*ptr=(char*)&i;
  printf("%d",*ptr);
  return 0;
 }
