A double-word (Image-1)

• Is a group of two consecutive Words
• A double-word combines 4 bytes, or 8 nibbles, or 32 bits

The bits

• Counted from right to left, start at 0 and end at 31 (Image-1)

(Image-2)

Considered as a group of 32 bits (Image-2)

• The most right bit, bit 0, is called the Low Order bit or LO bit or LOBIT
• The most left bit, bit 31, is called the High Order bit or HI bit or HIBIT
• The other bits are referred to using their positions: bit 1, bit 2, bit 3, etc

Considered as a group of 4 bytes (Image-2)

• The group of the first 8 bits (from bit 0 to bit 7), which is the right byte, is called the Low Order Byte, or LO Byte. It is sometimes referred to as LOBYTE
• The group of the last 8 bits (from bit 24 to bit 31), which is the left byte, is called the High Order Byte, or HI Byte or HIBYTE
• The other bytes are called by their positions

Considered as a group of 2 words (Image-2)

• The group of the right 16 bits, or the right Word, is called the Low Order Word, or LO Word, or LOWORD
• The group of the left 16 bits, or the left Word, is called the High Order Word, or HI Word, or HIWORD

The minimum binary number you can represent with a double-word

• Is 0

The minimum decimal value of a double-word

• Is 0

To find out the maximum decimal value of a double-word

• You can use the base 2 formula giving a 1 value to each bit (B-1):

Example of (B-1)

• 1*2³¹+1*2³⁰+1*2²⁹ + 1*2²⁸ + 1*2²⁷ + 1*2²⁶ + 1*2²⁵ + 1*2²⁴ + 1*2²³ + 1*2²² + 1*2²¹ + 1*2²⁰ + 1*2¹⁹ + 1*2¹⁸ + 1*2¹⁷ + 1*2¹⁶ + 1*2¹⁵ + 1*2¹⁴ + 1*2¹³ + 1*2¹² + 1*2¹¹ + 1*2¹⁰ + 1*2⁹ + 1*2⁸ + 1*2⁷ + 1*2⁶ + 1*2⁵ + 1*2⁴ + 1*2³ + 1*2² + 1*2¹ + 1*2⁰

  = 2,147,483,648 + 1,073,741,824 + 536,870,912 + 268,435,456 + 134,217,728 + 67,108,864 + 33,554,432 + 16,777,216 + 8,388,608 + 4,194,304 + 2,097,152 + 1,048,576 + 524,288 + 262,144 + 131,072 + 65,536 + 32,768 + 16,384 + 8,192 + 4,096 + 2,048 + 1,024 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1

  = 4,286,578,708

The minimum hexadecimal value you can store in a double-word

• Is 0×00000000000000000000000000000000 which is same as 0×0

To find out the maximum hexadecimal number you can represent with a word

• Replace every group of 4-bits with an f or F (Image-3)

(Image-3)

A word

• Is a group of 16 consecutive bits

The bits

• Are counted from right to left starting at 0 (Image-1):

Considered as a group of 16 bits (Image-1)

• The most right bit of a word, bit 0, is called the least significant bit or Low Order bit or LO bit or LOBIT
• The most left bit, bit 15, is called the most significant bit or High Order bit or HI bit or HIBIT
• The other bits are referred to using their positions: bit 1, bit 2, bit 3, etc

(Image-2)

Considering that a word is made of two bytes (Image-2)

• The group of the right 8 bits is called the least significant byte or Low Order byte or LO byte or LOBYTE
• The other group is called the most significant byte or High Order byte or HI byte or HIBYTE

The most fundamental representation of a word in binary format

• Is 0000000000000000 (B-1)

(B-1) To make it easier to read

• You can group bits in 4, like this: 0000 0000 0000 0000 (B-2)

(B-2) The minimum binary value represented by a word

• Is 0000 0000 0000 0000

(B-2) The maximum binary value represented by a word

• Is 1111 1111 1111 1111

The minimum decimal value of a word

• Is 0

To find out the maximum decimal value of a word

• You can use the base 2 formula, filling out each bit with 1 (B-3):

Example of (B-3)

• 1*2¹⁵+1*2¹⁴+1*2¹³ + 1*2¹² + 1*2¹¹ + 1*2¹⁰ + 1*2⁹ + 1*2⁸ + 1*2⁷ + 1*2⁶ + 1*2⁵ + 1*2⁴ + 1*2³ + 1*2² + 1*2¹ + 1*2⁰

= 32768 + 16384 + 8192 + 4096 + 2048 + 1024 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1
= 65535

The minimum hexadecimal value you can store in a word

• Is 0×0000000000000000 (B-4)

(B-4) This is also represented

• As 0×00000000, or 0×0000, or 0×0
• All these numbers produce the same value, which is 0×0

 To find out the maximum hexadecimal number you can store in a word

• Replace every group of 4 bits with an f or F (Image-3)

(Image-3)

Counting on base and extending the table we used earlier

• We would get (Image-4)

(Image-4)