Question

Consider the following code:

int i = 0x56789a;

char c = (char) i;

int j = c;

What is the hex value of c? (0x9a)

What is the hex value of j? (0xffffff9a)

I know the answer but I don't know how to get to that answer.

Can someone explain this to me, please? Thanks!

Answer 1

int i = 0x56789a;
char c = (char) i;

char can only store a single byte.
and when you cast an integer to a char. then copy right most byte(0x9a) from i into char c.
so, c becomes (0x9a)

int j = c;

Converting 9A to binary
9 => 1001
A => 1010
So, in binary 9A is 10011010

we know that size of int is 32 bits.
when we copy a char(8 bits) into int, then copy the left most bit of c into remaining bits of int

so, int in binary is 11111111 11111111 11111111 10011010
now, convert this to hexadecimal

Hexadecimal     Binary
    0           0000
    1           0001
    2           0010
    3           0011
    4           0100
    5           0101
    6           0110
    7           0111
    8           1000
    9           1001
    A           1010
    B           1011
    C           1100
    D           1101
    E           1110
    F           1111
Use this table to convert from binary to hexadecimal
Converting 11111111111111111111111110011010 to hexadecimal
1111 => F
1111 => F
1111 => F
1111 => F
1111 => F
1111 => F
1001 => 9
1010 => A
So, in hexadecimal 11111111111111111111111110011010 is 0xFFFFFF9A

that's why value of i contains value of 0xffffff9a