Char pointer to access an int

Question

main()
{
    int i =300;
    char *ptr=&i;
    *++ptr=2;
    printf("%d",i);
}

Answer 1

Here, we are modifying an integer variable using a char pointer. The modification happens to the second byte (from the left of the starting location) and it is changed to 2. 

i.e., 300 will be stored as (lower address on left) on a big endian machine

0 * 224

0 * 216

1 * 28

44 * 20

and as 

44 * 20

1 * 28

0 * 216

0 *224

on a little endian machine.

In both the cases using the char pointer we point to the starting byte. And we are incrementing the starting byte by 2. 

Ao, after the increment we have 

0 * 224

2 * 216

1 * 28

44 * 20

on big endian architectures and

44 * 20

2 * 28

0 * 216

0 *224

on little endian architectures. 

The respective values are 131372 and 556.

(This is the most common method of checking if an architecture is little endian or big endian)

Comments

char *ptr = &i;

Here we are using a char pointer to hold an address to integer, which many compiler won't like. The correct way is

char *ptr = (char *) &i;

Here, we explicitly tell the compiler that we want the integer address to be treated as an address to character which is legal in C.
(A character pointer can be used to address any other data type like above, but this is not always true for integer pointer due to alignment considerations.)

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Why is it that modification done to the second byte ?@ Arjun Sir

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because in the code we are telling compiler to do exactly that- character means 1 byte, when we increment a character pointer it points to second byte - whether the pointed value is an integer or whatever is not the concern and the only thing that matters is the type of pointer- which here is char.