07oct_2.c

#include <stdio.h>

int main()
{
    char arr[] = {'a', 'b', 'c', 'd', 'e'};
    int *ptr = arr;
    printf("%c", *(ptr + 1));
    return 0;
}

/*
-This C code snippet is designed to print the second character of a character array arr. 
-The array arr is declared and initialized with five characters: 'a', 'b', 'c', 'd', and 'e'. 
-A pointer ptr of type int is declared and assigned the address of the first element of the character array arr. 
-However, this is incorrect and can lead to undefined behavior because ptr is an int pointer, but it's assigned the address of a char array. 
-A char pointer should be used instead.

-The printf statement is used to print the second character of the array. 
-The expression ptr + 1 increments the pointer ptr by one element of its type (int). 
-Since ptr is an int pointer, incrementing it by one moves the pointer to the next int location, not the next char location. 
-This is incorrect and can lead to undefined behavior. The * operator dereferences the pointer, accessing the value at the incremented location. 
-The %c format specifier in printf expects a char argument, but the expression *(ptr + 1) evaluates to an int value due to the incorrect pointer type.

-To fix the issues in the original code, you should use a char pointer instead of an int pointer. 
-With this correction, the code will correctly print the second character 'b' of the array arr.
*/

/*
# The size of a pointer variable in C depends on the architecture of the system. In most modern systems, a pointer is typically 4 bytes (32 bits) or 8 bytes (64 bits) in size.

-       On a 32-bit system, the size of a pointer is usually 4 bytes (32 bits).
-       On a 64-bit system, the size of a pointer is usually 8 bytes (64 bits).
*/