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Data representation

Data representation. How do we represent data in a digital system? Digital data is stored as bits : a bit is a binary number {0, 1} in a hardware design this is usually interpreted as 0 Volt, GND {5, 3.3, 1.8, … } Volt, VCC. String of bits!.

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Data representation

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  1. Data representation • How do we represent data in a digital system? • Digital data is stored as bits: • a bit is a binary number {0, 1} • in a hardware design this is usually interpreted as • 0 Volt, GND • {5, 3.3, 1.8, … } Volt, VCC

  2. String of bits! Everything we work with is represented as a string of bits.Example: • We don't work with integers! • We represent (encode) integers as strings of bits and manipulate those. 1 0 1 0 (binary) represents 1*23 + 0*22 + 1*21 + 0*20 (decimal)8 + 0 + 2 + 0 = 10 (decimal)

  3. Ask yourself: • How many bits are required to represent this thing? • How can we do it? • We need to structure the information.

  4. Word, halfword and byte • “The word: 32 bits”. • “The halfword: 16 bits”. • “The byte: 8 bits”. 31 0 High order bit Low order bit 0 15 0 7

  5. “The Space” • We can group a large number of bit strings into an array called a space. • We use at least three spaces: • the register space. • the instruction space. • the data space. • Do not confuse them!

  6. The Register Space (or Register “file”) • An array of words. • Smallest addressable unit is the word. • 32 choices, how do we select? • 2x = 32, x? • x=5 • i.e., we need 5 bits! 0 1 2 3 29 30 31

  7. Address • To work with a space we must be able to address the elements of the space. • Each space may have a different address method.

  8. The Instruction space (or instruction memory) • An array of bytes. • Smallest addressable unit is a byte, butthe smallest accessible unit is a word. • 232 bytes gives 230 accessible words.

  9. What's that mean? • An instruction address is 32 bits long. • Same as a word. • An instruction address fits into a word!

  10. But: • Only every fourth address is legal. • The ones whose last two bits are “00”. • We can access 230 instructions.

  11. Instruction memory 1 0 1 2 3 • Address “4”: 0000........00100 • Hardware accesses an entire 32-bit word. 4 5 6 7 8 9 10 11 12 13 14 15

  12. Instruction memory 2 0 1 2 3 • Address "3": 0000........00011 • The hardware will complain! • This is called an “alignment error”. • The address was not “word aligned”. 4 5 6 7 8 9 10 11 12 13 14 15

  13. Byte order Byte 0 Byte 1 Byte 2 Byte 3 • Big endian • Little endian Addr 0 Addr 1 Addr 2 Addr 3 Byte 3 Byte 2 Byte 1 Byte 0 Addr 0 Addr 1 Addr 2 Addr 3

  14. Instruction memory 3 0 1 2 3 • An instruction is represented by 32 bits. • There are two “instruction formats”. (Not the whole truth, but don't worry about it) 4 5 6 7 8 9 10 11 12 13 14 15

  15. Instruction formats 6 bits 5 5 5 Remember: • Never immediate field together with rd-field. • Immediate field is only 16 bits! Format 1 (reg) Opcode rs rt rd 6 bits 5 5 16 Opcode Format 2 (imm) Immediate rs rt

  16. “Data space” • 232 bytes, byte addressable, byte accessible. • A data address fits into 32 bits. • Halfword accesses must be halfword aligned; (low order bit = 0). • Word accesses must be word aligned; (low order 2 bits = 0).

  17. Review - How many bits?? • Instruction address ? • Instruction ? • Immediate field ? • Register address ? • Register ? • Data address ? • Data ?

  18. Review - How many bits?? • Instruction address 32, word aligned • Instruction 32 • Immediate field 16 • Register address 5 • Register 32 • Data address 32, byte aligned • Data 8 (byte) 16 (halfword) 32 (word)

  19. Questions? Feedback from the lectures: Per.Lindgren@ltu.se

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