Computer Organization and Architecture Lecture 15

From Computer Organization and Architecture Lecture 14

---

IEEE 754 Floating-Point Representation

Introduction

IEEE 754 is the standard for representing floating-point numbers in computers. It ensures consistency across different computing systems and architectures. The standard defines multiple formats, including single precision (32-bit), double precision (64-bit), and extended precision formats.

Structure of IEEE 754 Floating-Point Number

A floating-point number in IEEE 754 format consists of three main components:

1. Sign Bit (S):

   • 1 bit that determines the sign of the number.
   • 00 for positive, 11 for negative.

2. Exponent (E):

   • Stored with a bias (e.g., 127 for single precision, 1023 for double precision).
   • Determines the scaling factor.

3. Mantissa (M) (Fraction):

   • Stores the significant digits of the number.
   • The actual significand is 1.M (implicit leading 1 for normalized numbers).

Mathematical Representation

A floating-point number in IEEE 754 format is represented as:

$\pm M\times B^E$

Where:

• MM is the mantissa (also called the significand).
• BB is the base, which is 2 in binary representation.
• EE is the exponent, calculated as:

$E=\text{Stored Exponent}-\text{BiasE}=\text{Stored Exponent}-\text{Bias}$

For single precision (32-bit):

$E=Exponent-127E=\text{Exponent}-127$

For double precision (64-bit):

$E=Exponent-1023E=\text{Exponent}-1023$

---

Questions

Convert → $1259.125_{10}$ into single & double pre format

Solve

• $(1259{)}_{10}\to (10011101011{)}_2$
• $(0.125{)}_{10}\to (001{)}_2$ $(10011101011.001{)}_2⟹1.00111010110001\times 2^{10}$ $E-127=10$ $E=130$ $E=(130{)}_{10}$ $E=(10000010{)}_2$
• $E=10+127=137\to (10001001{)}_2$
• $M=00111010110001000000000$
• $S=0$
• $01000100100111010110001000000000$

For Double Precision Compare with Equation $E-1023=10$ $E=1033=(100000010001{)}_2$

• $E=10+1023=1033\to (10000001001{)}_2$
• $M=00111010110001000000000$
• $S=0$

S	E	M

Question 2

$(263.3{)}_{10}$

• $(263{)}_{10}\to (100001111{)}_2$
• $(0.3{)}_{1}0\to (00100111\dots {)}_2$

Compare with Equation $100001111.01010101\dots$

$1.0000111101010101\cdots \times 2^8$

Comparing

$E=8+127$ $E=135$

$(E{)}_2=(10000111{)}_2$ $E$

S	E	M
0	0b10000111	…

Question 3

$-786.25$

Solution

• $(786{)}_{10}=(1100010010{)}_2$
• $(0.25{)}_{10}=(01{)}_2$
• $1100010010.01=1.10001001001\ast 10^9$
• $\pm (1.N)\times 2^{n+127}\text{ because Negative Bit }$
• E = 136

S	E	M
1	0b10001000	0b10001001001

Information

• date: 2025.03.06
• time: 16:11