CSI 333 Lecture 18 MIPS Assembly Language (MAL): Part VI (Representing Negative Integers) 18 1 / 17

Transcription

1 CSI 333 Lecture 18 MIPS Assembly Language (MAL): Part VI (Representing Negative Integers) 18 1 / 17

2 Representing Negative Integers Ref: Section 2.4 of Patterson & Hennessey text. Numbering of Bits: n 1 n... Big Endian MSB LSB n n Little Endian MSB Word size = n+1 bits LSB Notes: We will assume Little Endian bit numbering. The most significant bit is the sign bit (0 for non-negative, 1 for negative) / 17

3 Representing Negative Integers (continued) Known Methods: Sign Magnitude representation. (Not common) One s complement representation. (Not common) Two s complement representation. (Common) Biased representation. (Used in special situations) 18 3 / 17

4 Sign Magnitude Form Example: (with word size n + 1 = 8 bits) represents represents -40 Range: For a word size of n + 1, Smallest integer = (2 n 1) Largest integer = +(2 n 1) Example: With n + 1 = 8, the range of integers representable using sign-magnitude form is 127 through / 17

5 Sign Magnitude Form (continued) Formula for value: Bit string : b n b n 1... b 1 b 0 n 1 Value : ( 1) bn b i 2 i i=0 Disadvantage: The integer zero has two representations: represents represents / 17

6 One s Complement Form Positive integers: sign bit = 0. Usual binary representation with Negative integers: Complement the binary representation of the corresponding positive value. (The sign bit is also complemented.) Example: represents represents -15 Problem: What decimal value does the 1 s complement binary number represent? Solution: When we complement all the bits, we get , which represents +109 decimal. So, the given integer = / 17

7 One s Complement Form (continued) Range: For a word size of n + 1, Smallest integer = (2 n 1) Largest integer = +(2 n 1) Example: With n + 1 = 8, the range of integers representable using 1 s complement form is 127 through Formula for value: Disadvantage: Bit string : b n b n 1... b 1 b 0 Value : n 1 b i 2 i b n (2 n 1) i=0 The integer zero has two representations: represents represents / 17

8 Two s Complement Form Positive integers: Usual binary representation, with sign bit = 0. Negative integers: Take 1 s complement and then add 1. Recall: Binary addition table. Inputs Sum Carry / 17

9 Two s Complement Form (continued) Problem: Find the 2 s complement representation of 15 using 8 bits. Solution: +15: s compl.: Add 1 : <-- Answer 18 9 / 17

10 Two s Complement Form (continued) Problem: Solution: Find the 2 s complement representation of 64 using 8 bits. +64: s compl.: Add 1 : <-- Carries <-- Answer Another method for 2 s complement: 1 Start with the binary representation of the positive value. 2 Copy bits from right to left, until the first 1 has been copied. 3 Complement every bit thereafter / 17

11 Two s Complement Form (continued) Problem: Solution: Find the 2 s complement representation of 64 using 8 bits. +64: s complement representation for -64: ^ (First 1 copied) Problem: What decimal value does the 2 s complement binary number represent? Solution: When we take the 2 s complement of the given number, we get , which represents +1 decimal. So, the given integer = 1 decimal / 17

12 Two s Complement Form (continued) Problem: What is the 2 s complement of ? Solution: Given: s compl.: Add 1 : <-- Carries <-- Answer (value = 0) 1 --> Carry out of sign bit (must be ignored) So, zero has a unique representation in 2 s complement form. Note: In 2 s complement arithmetic, the carry out of the sign bit is called the end around carry. It should be ignored. (It does not indicate overflow.) / 17

13 Two s Complement Form (continued) Range: For a word size of n + 1, Smallest integer = 2 n Largest integer = +(2 n 1) For example, with n + 1 = 8, the range of integers representable using 2 s complement form is 128 through (The range is asymmetric.) Formula for value: Bit string : b n b n 1... b 1 b 0 Value : n 1 b i 2 i b n 2 n i= / 17

14 Two s Complement Form (continued) A (minor) disadvantage: Because of the asymmetry of the range, we cannot take the 2 s complement of the smallest (i.e., most negative) value. Example: The 2 s complement representation for 128 decimal using 8 bits is (Verify this.) When we take the 2 s complement of , we get itself, which is incorrect. Reason: The expected result, namely +128, cannot be represented using 8 bits (including the sign bit) / 17

15 Biased Representation Used only in special situations. No sign bit (i.e., the resulting representation is unsigned). A suitable positive integer B is chosen as the bias. Integer i (positive or negative) is represented by the unsigned binary representation of the value B + i. Examples: Consider 8-bit representations with bias = Biased-127 representation of the decimal value +9 is the 8-bit unsigned representation of the integer = 136, namely Biased-127 representation of the decimal value 21 is the 8-bit unsigned representation of the integer = 106, namely / 17

16 Biased Representation (continued) Range: For a word size of n + 1 and bias B Smallest integer = B Largest integer = +(2 n+1 1 B) Example: With n + 1 = 8 and bias B = 127, the range of integers representable using Biased-127 form is 127 through (The range is asymmetric.) Formula for value: Bit string : b n b n 1... b 1 b 0 Bias : Value : B n b i 2 i B i= / 17

17 Biased Representation (continued) Notes: Used to store the exponents in the IEEE Floating Point Standard (IEEE FPS) representation. For a word size of n + 1 bits, the normally chosen bias values are either 2 n or 2 n 1. (Reason: This keeps the range of values reasonably symmetric around zero.) / 17