走進 機器層級程式觀點，在這裡硬體的複雜性被抽象為一個結構化的 虛擬地址空間。在此觀點下，記憶體並非變數的集合，而是一個龐大的、連續的 8 位元區塊陣列，稱為 字節。對於具有 $w$ 位元字長的機器而言，這些位址範圍從 $0$ 到 $2^w - 1$，定義了程式的可達範圍。
1. 十六的威力
二進位是電路的語言，但 十六進位表示法 是程式設計師的語言。因為 $16 = 2^4$，單一的十六進位數字（0–F）能完美對應到 4 位元的半字節。這使得一個 1 字節的值可以用恰好兩個數字來簡潔地表示（例如， 0xFF）。這種簡寫對於閱讀 機器碼 和 組合語言碼至關重要，例如指令 4004dc: 48 03 47 28。
2. 精確度與運算
當我們操作 整數資料類型時，會遇到 布林環 和 二補數 邏輯。我們必須處理 小端序 儲存， 整數溢位以及 單精度 浮點數中 無限大 ($+\infty$) 和 NaN 存在。理解這些位元模式是掌握 任意大小的運算 及穩健的系統程式設計的第一步。
main.py
TERMINALbash — 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
✅ Correct!
0x3=0011, 0x9=1001, 0xA=1010, 0x7=0111. For the second: 1101=D, 1010=A, 0111=7, 1101=D.❌ Incorrect
Each hex digit corresponds to a 4-bit nibble. Match 0x3 to 0011, 0x9 to 1001, etc.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
✅ Correct!
0x55 (01010101) and 0x46 (01000110). AND results in 01000100 (0x44). XOR results in 00010011 (0x13).❌ Incorrect
Convert hex to binary, apply the bitwise operator bit-by-bit, and convert back to hex.QUESTION 3
In a 4-bit Two's-Complement system, what is the unsigned representation (T2U) of the value -1?
1
15
7
8
✅ Correct!
$T2U_w(x) = x + 2^w$ for $x < 0$. So $-1 + 16 = 15$.❌ Incorrect
A signed -1 in 4-bit binary is 1111, which represents 15 in unsigned.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)
✅ Correct!
x XOR y is 0 if and only if x equals y. The logical NOT (!) then turns 0 into 1 (True).❌ Incorrect
The XOR operator returns 0 for identical bits. How do we turn a 0 into a logical true?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
✅ Correct!
0xC (12) + 0x8 (8) = 20 (0x14). So 0x503c + 8 = 0x5044. 0x3c (60) - 0x40 (64) = -4, borrow from next digit: 0x4FFC.❌ Incorrect
Remember that hex carries happen at 16, not 10. $C + 8 = 14_{16}$.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.
Q
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.
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.
Q
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.
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.
Q
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.
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.