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- 1. Chapter 2 Instructions: Language of the Computer
- 2. Data Representation Integers A decimal example 2734 = 2 * 103 + 7 * 102 + 3 * 101 + 4 * 100 A binary example [01011]2 = 0 * 24 + 1 * 23 + 0 * 22 + 1 * 21 + 1 * 20 = [11] 10 Unsigned Unsigned integers are positive (or to be more precise, do not have a sign) Using 32 bits 0 to +4,294,967,295 Chapter 2 Instructions: Language of the Computer 2
- 3. Data Representation n bits 0, (2n 1) For n = 4 use hexadecimal So for example: 0000 0000 0000 0000 0000 0000 0000 1011 becomes 0x0000000B Chapter 2 Instructions: Language of the Computer 3
- 4. Data Representation Signed Positive numbers are like unsigned But how do we store negative numbers? Chapter 2 Instructions: Language of the Computer 4
- 5. How do we store negative numbers? Lets represent -23: Start by representing the absolute value which is 23 in this case (For now use an 8-bit number instead of a 32-bit number to get rid of all the leading zeros) Chapter 2 Instructions: Language of the Computer 5
- 6. Negative Numbers Could leave leftmost bit for sign 0 for positive, 1 for negative Lead to problems Two zeros (0 and -0) which can be a big problem Could create infinite loops Could set leftmost bit to 1 i.e. -27 = -128 Then ADD as many 1s so that the sum of their decimal value and -128 equals the negative number we are representing We can get the same result with a better algorithm Note: that -1 is represented by 1111 1111 Chapter 2 Instructions: Language of the Computer 6
- 7. 2s Complement Find the complement of a binary number by flipping all digits (replacing 0s and 1s) Signed Negation Example: negate +2 +2 = 0000 0000 00102 2 = 1111 1111 11012 + 1 2s Complement Best Notation to use = 1111 1111 11102 Chapter 2 Instructions: Language of the Computer 7
- 8. 2s-Complement Signed Integers Given an n-bit number n 1 n 2 1 0 x x n 12 xn 2 2 x12 x0 2 Range: 2n 1 to +2n 1 1 Example 1111 1111 1111 1111 1111 1111 1111 11002 = 1231 + 1230 + + 122 +021 +020 = 2,147,483,648 + 2,147,483,644 = 410 Using 32 bits 2,147,483,648 to +2,147,483,647 Chapter 2 Instructions: Language of the Computer 8
- 9. 2s-Complement Signed Integers Bit 31 is sign bit 1 for negative numbers 0 for non-negative numbers (2n 1) cant be represented Non-negative numbers have the same unsigned and 2s-complement representation Some specific numbers 0: 0000 0000 0000 1: 1111 1111 1111 Most-negative: 1000 0000 0000 Most-positive: 0111 1111 1111 Chapter 2 Instructions: Language of the Computer 9
- 10. Sign Extension Representing a number using more bits Preserve the numeric value In MIPS instruction set addi: extend immediate value lb, lh: extend loaded byte/halfword beq, bne: extend the displacement Replicate the sign bit to the left c.f. unsigned values: extend with 0s Examples: 8-bit to 16-bit +2: 0000 0010 => 0000 0000 0000 0010 2: 1111 1110 => 1111 1111 1111 1110 Chapter 2 Instructions: Language of the Computer 10
- 11. Data Representation We need a way to map: data binary Data Types Number Integer Signed/Unsigned Real Char Other (Picture, etc.) How .byte 11 (in MIPS) is stored in memory: 0000 0000 0000 0000 0000 0000 0000 1011 Chapter 2 Instructions: Language of the Computer 11
- 12. Byte Ordering (Endianess) How are bytes ordered (numbered) within a word? b y te o f f se t little endian byte 0 lsb of word msb 3 2 1 0 0 1 2 3 msb lsb 0 1 2 3 big endian byte 0 b y te o f f se t of word Little Endianaddress (item) address (least significant byte) ; Intel 80x86, DEC Alpha, etc. Big Endian address (item) address (most significant byte) ; HP PA, IBM/Motorola PowerPC, SGI (MIPS), Ultra Sparc, etc. Significant when binary data (int, float, double, etc.) need to be transferred from one machine to another. Internet uses the Big Endian byte order. Chapter 2 Instructions: Language of the Computer 12
- 13. Memory Operands Main memory used for composite data Arrays, structures, dynamic data To apply arithmetic operations Load values from memory into registers Store result from register to memory Memory is byte addressed Each address identifies an 8-bit byte Words are aligned in memory Address must be a multiple of 4 MIPS is Big Endian Most-significant byte at least address of a word c.f. Little Endian: least-significant byte at least address Chapter 2 Instructions: Language of the Computer 13
- 14. Memory Operand Example 1 C code: g = h + A[8]; g in $s1, h in $s2, base address of A in $s3 Compiled MIPS code: Index 8 requires offset of 32 4 bytes per word lw $t0, 32($s3) # load word add $s1, $s2, $t0 offset base register Chapter 2 Instructions: Language of the Computer 14
- 15. Memory Operand Example 2 C code: A[12] = h + A[8]; h in $s2, base address of A in $s3 Compiled MIPS code: Index 8 requires offset of 32 lw $t0, 32($s3) # load word add $t0, $s2, $t0 sw $t0, 48($s3) # store word Chapter 2 Instructions: Language of the Computer 15
- 16. Registers vs. Memory Registers are faster to access than memory Operating on memory data requires loads and stores More instructions to be executed Compiler must use registers for variables as much as possible Only spill to memory for less frequently used variables Register optimization is important! Chapter 2 Instructions: Language of the Computer 16
- 17. Immediate Operands Constant data specified in an instruction addi $s3, $s3, 4 No subtract immediate instruction Just use a negative constant addi $s2, $s1, -1 Design Principle 3: Make the common case fast Small constants are common Immediate operand avoids a load instruction Chapter 2 Instructions: Language of the Computer 17

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