Question

Given byte-addressed memory with location 0x100 containing 0x62 and location 0x101 containing 0x7C. What is the decimal value of the signed short stored at location 0x100:

1. Assuming big-endian addressing.

2. Assuming little-endian addressing.

You may assume that a short occupies 2 bytes.

Answer 1

a)
big-endian
hex value is 0x627C

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 hexadecimal to binary
Converting 627C to binary
6 => 0110
2 => 0010
7 => 0111
C => 1100
So, in binary 627C is 0110001001111100

Now converting 0x627C to decimal
since left most bit is 0, this number is positive
so, we can directly convert this into a decimal value
Converting 110001001111100 to decimal
110001001111100
=> 1x2^14+1x2^13+0x2^12+0x2^11+0x2^10+1x2^9+0x2^8+0x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+0x2^1+0x2^0
=> 1x16384+1x8192+0x4096+0x2048+0x1024+1x512+0x256+0x128+1x64+1x32+1x16+1x8+1x4+0x2+0x1
=> 16384+8192+0+0+0+512+0+0+64+32+16+8+4+0+0
=> 25212
Answer: 25212

b)
little-endian
hex value is 0x7C62

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 hexadecimal to binary
Converting 7C62 to binary
7 => 0111
C => 1100
6 => 0110
2 => 0010
So, in binary 7C62 is 0111110001100010

Now converting 0x7C62 to decimal
since left most bit is 0, this number is positive
so, we can directly convert this into a decimal value
Converting 111110001100010 to decimal
111110001100010
=> 1x2^14+1x2^13+1x2^12+1x2^11+1x2^10+0x2^9+0x2^8+0x2^7+1x2^6+1x2^5+0x2^4+0x2^3+0x2^2+1x2^1+0x2^0
=> 1x16384+1x8192+1x4096+1x2048+1x1024+0x512+0x256+0x128+1x64+1x32+0x16+0x8+0x4+1x2+0x1
=> 16384+8192+4096+2048+1024+0+0+0+64+32+0+0+0+2+0
=> 31842
Answer: 31842