Review

• Chapter 2: secs. 2.1, 2.2.1, 2.2.2, 2.2.3
• Notions of bits and bytes. (1 byte = 8 bits)
• Integral types
• Relevant operations in C for this class: bit-level operations (&, |, ^,~), logical operations (&&, ||, !), shift (left and right)

Index Card Questions

• Why is !0x00 == 0x01 and not 0x11?
  • ! is a logical operator
  • applying ! to 0 (false) is 1 (true)
  • applying ~ to 0 (assuming 32 bits) is FFFFFFFF in hex
• How can you tell if a binary vector is meant to represent a number or a subset?
• How to differentiate between bitwise AND (&) and the operator & (as give me the address of)?
• Does C have a NAND operator?

Overview

• Use of shifts and masks
• Encoding integers

Practice Masking Operations in C

• Problem 2.12 p. 55: Write C expressions for
  • Least significant byte of x, with all other bits set to 0.
  • All but the least significant byte of x complemented, with least significant byte left unchanged.
  • Least significant byte set to all ones, and all other bytes of x left unchanged.
• Do for x = 0x789ABC21 and w = 32 (any w >= 8)

Extracting and Assembling Integral Types - Shift and Mask

• Shift/AND commonly used to extract subset of bits from integral types

  int x, y, z;
  // Pull bits 16-23 out of X
  y = (x >> 16) & 0xff;

• OR and shift used to assemble multiple values into a single larger value

  // Construct z out of those bits and another value;
  z = (y & 0xff) | ((0x18 << 8) & 0x3f00);

• Used in places where you pack multiple values into a single word (e.g. device drivers)

• C unions, structs, and bit-fields are another way to do this (more later)

Problems to Work Through

• Problems: (no answer in the book)
  • 2.59,
  • 2.61,
  • 2.68
    
• With coding rules on pp. 128-129
• Homework #1 will be posted tonight, includes some of these and some others.

Learning Objectives

• After this session + practice you should be able to:
  • Describe how are integral types represented in the computer: unsigned and signed.
  • Define and perform conversion, casting, expanding and truncating operations on integers.
  • Define and perform integer arithmetic operations + Explain and analyze the errors derived from these operations.

Review: 1s Complement

Review: 1s Complement Problems

Review: 1s Complement Table

3-bit one’s-complement:

Review: 2s Complement

Review: 2s Complement Benefits

Review: 2s Complement Table

3-bit two’s-complement:

Encoding Integers

• Converting base 2 to base 10 (X = binary number, w = word size):

  • Unsigned: $UB2D(X) = \sum_{i=0}^{w-1} x_i * 2^i$
  • 2s complement: $TB2D(X) = -x_{w-1} * 2^{w-1} + \sum_{i=0}^{w-2} x_i * 2^i$

• C short is 2 bytes long

  	Decimal	Hex	Binary
  x	15213	3B 6D	00111011 01101101
  y	-15213	C4 93	11000100 10010011

Try It

• Show how to represent 13₁₀ as an 8 bit value, MSB is the sign bit.
• Then show how to represent −13₁₀ in 8 bit 2’s complement notation.
• Finally show that 13₁₀ + −13₁₀ = 0 using the 2’s complement representations.

Try It

• Show how to represent 13₁₀ as an 8 bit value, MSB is the sign bit.
  • 00001101 = 1 * 2³ + 1 * 2² + 1 * 2⁰ = 13₁₀
• Then show how to represent −13₁₀ in 8 bit 2’s complement notation.
• Finally show that 13₁₀ + −13₁₀ = 0 using the 2’s complement representations.

Try It

• Show how to represent 13₁₀ as an 8 bit value, MSB is the sign bit.
  • 00001101 = 1 * 2³ + 1 * 2² + 1 * 2⁰ = 13₁₀
• Then show how to represent −13₁₀ in 8 bit 2’s complement notation.
  • 11110011 = −1 * 2⁷ + 1 * 2⁶ + 1 * 2⁵ + 1 * 2⁴ + 1 * 2¹ + 1 * 2⁰ = −13₁₀
• Finally show that 13₁₀ + −13₁₀ = 0 using the 2’s complement representations.

Try It

• Show how to represent 13₁₀ as an 8 bit value, MSB is the sign bit.
  • 00001101 = 1 * 2³ + 1 * 2² + 1 * 2⁰ = 13₁₀
• Then show how to represent −13₁₀ in 8 bit 2’s complement notation.
  • 11110011 = −1 * 2⁷ + 1 * 2⁶ + 1 * 2⁵ + 1 * 2⁴ + 1 * 2¹ + 1 * 2⁰ = −13₁₀
• Finally show that 13₁₀ + −13₁₀ = 0 using the 2’s complement representations.
  • 00001101 + 11110011 = 00000000

-13 2’s Complement Step By Step

Converting −13₁₀ to 2’s complement:

1. Convert absolute value to binary: 13₁₀ = 00001101

2. Invert all the bits: 11110010

3. Add 1: 11110011

4. Verify, recall $TB2D(X) = -x_{w-1} * 2^{w-1} + \sum_{i=0}^{w-2} x_i * 2^i$:

   128	64	32	16	8	4	2	1
   1	1	1	1	0	0	1	1
   -128	64	32	16	0	0	2	1	= −1310

More Practice

• Represent 20₁₀ as an 8 bit value, MSB is the sign bit.
• Represent −20₁₀ as an 8 bit 2’s complement value, MSB is the sign bit.

Numeric Ranges

Implications For C Programming

• Many of these are constants in C:
• #include<limits.h>
• Declares constants such as:
  • ULONG_MAX
  • LONG_MAX
  • LONG_MIN