C Programming: Understanding Array Initialization and Pointer Arithmetic in int main()

There are a few issues with this code:

1. The initialization of the array 'a' is incorrect. In C, the values of an array must be enclosed in curly braces '{}'. So it should be: 'int a[5] = {1,2,3,4,5};'

2. The expression '(&a + 1)' is taking the address of the array 'a', adding '1' to it (which advances the pointer by the size of the array, which is '5*sizeof(int)'), and then casting it to a pointer to 'int'. This results in 'p' pointing to some memory location beyond the end of the array 'a', which is undefined behavior.

3. The 'printf' statement is trying to print the second element of the array 'a' and the last element of the memory block pointed to by 'p'. But since 'p' points to undefined memory, this could result in a segmentation fault or other undefined behavior.

Here's a corrected version of the code:

#include <stdio.h>

int main()
{
    int a[5] = {1, 2, 3, 4, 5};
    int *p = &a[0]; // point to the first element of a
    p++; // advance the pointer by one element
    printf("first_data = %d, second data = %d\n", a[1], *(p-1));
    return 0;
}

This code initializes the array 'a' correctly, sets 'p' to point to the second element of 'a' (by incrementing the pointer by one element), and then prints the values of the second element of 'a' and the element immediately before 'p'. The output should be:

first_data = 2, second data = 1