Computer Systems: A Programmer's Perspective (Global Edition): Hexadecimal Notation and the Virtual Address Space

Step into the Machine-Level Program View, where the complexity of hardware is abstracted into a structured Virtual Address Space. In this view, memory is not a collection of variables, but a massive, contiguous array of 8-bit blocks called bytes. For a machine with a $w$-bit word size, these addresses range from $0$ to $2^w - 1$, defining the limits of the program's reach.

1. The Power of 16

Binary is the language of circuits, but Hexadecimal Notation is the language of programmers. Because $16 = 2^4$, a single hex digit (0–F) maps perfectly to a 4-bit nibble. This allows a 1-byte value to be expressed compactly by exactly two digits (e.g., 0xFF). This shorthand is essential for reading machine code and assembly code, such as the instruction 4004dc: 48 03 47 28.

2. Precision and Arithmetic

As we manipulate Integral Data Types, we encounter Boolean rings and Two's-complement logic. We must navigate Little endian storage, Integer overflow, and the nuances of single precision floating point where Infinity ($+\infty$) and NaN reside. Understanding these bit patterns is the first step in mastering Arbitrary size arithmetic and robust systems programming.

TERMINAL bash — 80x24

> Ready. Click "Run" to execute.

QUESTION 1

Perform the following number conversions: 0x39A7 to binary, and binary 1101101001111101 to hexadecimal.

0x39A7 = 0011100110100111; Binary = 0xDA7D

0x39A7 = 1111001110100111; Binary = 0xDA7D

0x39A7 = 0011100110100111; Binary = 0xEA7D

0x39A7 = 0011100111100111; Binary = 0xDA8D

QUESTION 2

Suppose a=0x55 and b=0x46. What is the result of the bitwise operations 'a & b' and 'a ^ b' in hexadecimal?

a & b = 0x44; a ^ b = 0x11

a & b = 0x44; a ^ b = 0x13

a & b = 0x54; a ^ b = 0x13

a & b = 0x44; a | b = 0x55

QUESTION 3

In a 4-bit Two's-Complement system, what is the unsigned representation (T2U) of the value -1?

QUESTION 4

Which C expression is equivalent to (x == y) using only bit-level and logical operations?

!(x ^ y)

x & y

~(x | y)

!!(x ^ y)

QUESTION 5

What is the result of '0x503c + 0x8' and '0x503c - 0x40' in hexadecimal?

0x5044 and 0x4FFC

0x5040 and 0x4F9C

0x5044 and 0x5000

0x5034 and 0x4FFC

Case Study: Numerical Representation and Logic Errors

Analysis of Integer vs. Floating-Point Bit Patterns

The integer 3,510,593 has hexadecimal representation 0x00359141. However, the single-precision floating-point number 3,510,593.0 is represented as 0x4A546504. You must analyze why these differ and how computer arithmetic handles these structures.

1. Using the IEEE 754 formula $V = 2^E \times M$, explain why the floating-point hex 0x4A546504 corresponds to 3,510,593.0.

Solution:
The hex 0x4A546504 converts to binary: 0 [Sign] 10010100 [Exp] 10101000110010100000100 [Frac]. The exponent $e=148$, so $E = 148 - 127 = 21$. The significand $M = 1.frac$. Thus $V = 1.10101000110010100000100_2 \times 2^{21}$. Shifting the binary point 21 places right gives 1101010001100101000001.00_2, which is 3,510,593 decimal.

2. Identify the correlation between the bits of the integer (0x00359141) and the floating-point fraction bits.

Solution:
The integer 3,510,593 is $1101010001100101000001_2$. Notice that the significand of the float (ignoring the implicit leading 1) contains exactly these same bits shifted appropriately. The float bits are essentially a normalized version of the integer's bit pattern, shifted to fit the IEEE format.

3. Evaluate the C expression: 'x == $(int)(float)$ x' where x is a large 32-bit integer. Is it always true?

Solution:
False. A 32-bit 'int' can have up to 31 bits of precision. However, a single-precision 'float' only has 23 bits in its fraction (the significand). Therefore, large integers that require more than 23 bits of precision (like 3,510,593, which needs 22, but larger ones) will experience rounding errors when cast to float, losing their least significant bits.