Same C pointer showing different values? - c

Question

I'm running this program:
#include<stdio.h>
void main(){
int num = 1025;
int *poinTer = &num;
char *pointChar = poinTer+1;
*pointChar = 'A';
printf("Size of Integer: %d\n", sizeof(int));
printf("Address: %d, Value: %d\n", poinTer, *poinTer);
printf("Address: %d, Value: %c\n", poinTer+1, *(poinTer+1));
printf("Address: %d, Value: %c\n", pointChar, *pointChar);
}
*pointChar and *(poinTer+1) should output same result but the output that I'm getting is different. *pointChar is not outputting any value:
Size of Integer: 4
Address: 1704004844, Value: 1025
Address: 1704004848, Value: A
Address: 1704004673, Value:
What's happening here?

Answer 1

When you perform + 1 on a pointer, it does not necessarily increase the memory address by 1. It increases it by sizeof(*ptr).
In this case, poinTer + 1 is equivalent to (char*)poinTer + sizeof(int). This actually makes dealing with arrays much easier.
The good old fashioned ptr[i] is syntactic sugar for *(ptr + i). So, if you have an array of 10 integers, ptr[4] will point to the 5th element rather than 4 bytes away from the base address (since integers are generally 4 or 8 bytes).
So what you've actually done is:
Create an int (num) on the stack and gave it the value 1025
Created a int*(poinTer) on the stack and assigned it the memory address of num
Incremented the pointer by sizeof(int) (which unintentionally points to a different memory address), then cast it to a char* and assigned it to a new pointer.
Assigned the byte pointed to at this new memory address the value 65 ('A').
This is probably what you wanted to do:
#include<stdio.h>
void main(){
int num = 1025;
int *poinTer = &num;
char *pointChar = (char*)poinTer + 1;
*pointChar = 'A';
printf("Size of Integer: %d\n", sizeof(int));
printf("Address: %d, Value: %d\n", poinTer, *poinTer);
printf("Address: %d, Value: %c\n", (char*)poinTer + 1, *((char*)poinTer+1));
printf("Address: %d, Value: %c\n", pointChar, *pointChar);
}