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Variable and Data Types

Representing a Double-Word

(Image-1)

Double-word



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)

 A double-word is a group of two consecutive Words



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)

The maximum hexadecimal number you can store in a double-word


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