京都大学 情報学研究科 通信情報システム専攻 2020年8月実施 専門基礎A [A-7]

标签：

• Kyoto-University
• Number-Representation

Author

SUN

Description

Answer all the following questions.

(1) Answer the following questions on the binary number representation.
  (a) Express the following decimal numbers in the 8-bit two’s complement representation.
    (i) +88     (ii) −72

  (b) Convert the following 8-bit two’s complement binary numbers into the 8-bit sign-and-magnitude binary representation.
    (i) 10101010     (ii) 11010101

  (c) Show the results of the following additions and subtractions in the 8-bit sign-and-magnitude binary number system.
    (i) 00101010 + 10101011     (ii) 10101010 + 10101010
    (iii) 11010101 − 01010101     (iv) 10101011 − 10101010

  (d) Show the result of the following multiplication of 8-bit two’s complement binary numbers in the 16-bit two’s complement binary representation.
    01110111 × 10001000

(2) Explain the procedure of addition of floating-point numbers

(3) Explain ‘data addressing modes’ in processors.

题目描述

回答全部问题。

1. 关于二进制数表示：
   1. 将下列十进制数写成 8 位二进制补码： (i) $+88$；(ii) $-72$。
   2. 将下列 8 位补码转换为 8 位原码（符号—数值表示）： (i) 10101010；(ii) 11010101。
   3. 在 8 位原码体系中给出下列加减运算结果： (i) 00101010 + 10101011； (ii) 10101010 + 10101010； (iii) 11010101 − 01010101； (iv) 10101011 − 10101010。
   4. 将两个 8 位补码数 01110111 × 10001000 的乘积写成 16 位补码。
2. 说明浮点数加法的步骤。
3. 说明处理器中的“数据寻址方式”。

Kai

(1)

(a)

• (i) +88: $88 = 64 + 16 + 8 \rightarrow$ 01011000
• (ii) −72: $72 = 01001000$. Two’s complement: invert $\rightarrow$ 10110111, add 1 $\rightarrow$ 10111000

(b)

• (i) 10101010: Negative. Magnitude: invert $\rightarrow$ 01010101, add 1 $\rightarrow$ 01010110 (86). Sign-and-magnitude $\rightarrow$ 11010110
• (ii) 11010101: Negative. Magnitude: invert $\rightarrow$ 00101010, add 1 $\rightarrow$ 00101011 (43). Sign-and-magnitude $\rightarrow$ 10101011

(c)

• (i) 00101010 + 10101011: $+42 + (-43) = -1 \rightarrow$ 10000001
• (ii) 10101010 + 10101010: $-42 + (-42) = -84 \rightarrow$ 11010100
• (iii) 11010101 − 01010101: $-85 - (+85) = -170$. Since $|-170| > 127$, it results in overflow.
• (iv) 10101011 − 10101010: $-43 - (-42) = -1 \rightarrow$ 10000001

(d)

Multiply 01110111 (+119) and 10001000 (-120):

• Product: $119 \times (-120) = -14280$
• Magnitude: $14280 = 0011011111001000$
• Two’s complement for -14280: invert $\rightarrow$ 1100100000110111, add 1 $\rightarrow$ 1100100000111000

Result: 1100100000111000

(2)

1. Unpack: Extract sign, exponent, and significand.
2. Align: Shift significand of the smaller-exponent number until exponents match.
3. Add/Subtract: Perform operation on significands based on signs.
4. Normalize: Shift result to standard form and adjust exponent.
5. Round: Round the significand to the available precision.
6. Check: Verify exceptions (overflow, underflow, zero).

(3)

Addressing modes define how instructions locate operands:

• Immediate/Implied: Operand is constant within the instruction or implied by the opcode.
• Register: Operand is stored in a CPU register.
• Direct/Indirect: Address is directly in instruction, or instruction points to a pointer in memory.
• Register Indirect: A register holds the memory address (pointer).
• Indexed/Offset: Effective address = Base + Index or Base + Constant offset (common for arrays/stacks).
• PC-relative: Address is an offset from the Program Counter (used for branches).
• Auto-inc/dec: Register indirect with automatic address update before/after access.