Why does the outputs be "41 and 42" for this code in C?

Question

#include <stdio.h>

int main() {
int arr[3]={40,41,42};
int *ptr = (int*)(&arr +1);
printf("%d %d", *(arr +1), *(ptr-1));
return 0;
}

Answer 1

The first argument to printf is equivalent to arr[1] which prints the 2nd element of the array, which is 41.

arr is of type int (*)[3].

So *ptr = (int *)(&arr + 1) offsets the pointer by 1 * sizeof(int) * 3 and you end up on the first block past the end of the array.

But in the printf function you decrease the pointer offset by a single element *(p - 1) which gets the value of the last array element of the previous block which is 42.

You can think of it as:
int (*ptr)[3] = &arr;
ptr++;
printf("%d\n", (*ptr)[-1]);

where ptr++ jumps 3 integers ahead for each increment.

Answer 2

*(arr+1) is easy: adding 1 to arr (which is a pointer to an int) you will point to the second element of arr[], so *(arr+1) will be equivalent to arr[1]

*(ptr-1) is trickier: ptr is obtained adding 1 to &arr (not arr), which is a pointer to 3 int; this means that &arr+1 is equivalent to arr+3: it will point after the end of arr[].
If we understand this, it is easy to understand ptr-1: it is equivalent to arr+3-1, that is the last element of arr, arr[2]

Answer 3

Try:

int *ptr = (int*)(arr + 1);
.

Your code has pointer arithmetic problem.
&arr is of type int** hence the arithmetic is different from int*

Answer 4

See this code to get an idea of pointer arithmetic

https://code.sololearn.com/ckTvXAXVns6m/?ref=app

Answer 5

u added one to ur pointer address, so the behavior is undefined