#include<stdio.h>
int main(){
int i=30;
int *ptr=&i;
printf("Value of i is %d\n",*ptr);
ptr++;
printf("Value of i is %d\n",*ptr);
return 0;
}
//OUTPUT:-
Value of i is 30
Value of i is 6618648
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dbush
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Nirbhay Thakkar
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How do you know it is the address of the integer? Anyway, the code is invalid because the result of incrementing and accessing a pointer beyond the bounds of the variable it points to is undefined behaviour. – kaylum Oct 10 '21 at 21:41
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read the dupes. Consider `i` as a single element integer array. – 0___________ Oct 10 '21 at 21:44
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It is `(*ptr)++`, not what you wrote. – FaranAiki Oct 11 '21 at 02:38