Question

Consider three registers R1, R2 and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.

If R3 \(= \frac{{R1}}{{R2}},\) what is the value stored in R3?

1. 0x40800000
2. 0xC0800000
3. 0x83400000
4. 0xC85800000

Answer 1

Correct Answer - Option 2 : 0xC0800000

Concept:

 In IEEE- 754 single precision format, a floating-point number is represented in 32 bits.

Sign bit (MSB)

Biased Exponent (E’) (8 bits)

Normalized Mantissa (M’) (23 bits)

 

Sign bit value 0 means positive number, and 1 means a negative number.

The floating-point number can be obtained by formula: ± 1. M × 2^(E-127)

Data:

Content of R1: 0x 42200000               (0x means Hexadecimal notation)

Content of R2: 0x C1200000

Calculation:

Content of R1 in Hex (0x) is 42200000. After converting into binary, it can be represented in IEEE- 754 format as:

0

100 0010 0

010 0000 0000 0000 0000 0000

 

Sign bit is 0 i.e. the number is positive

Biased Exponent (E’) = 100 0010 0 = 132

Normalized Mantissa (M) = 010 0000 0000 0000 0000 0000 = .25

Therefore, the number in register R1 = + 1.25 * 2^(132-127) = 1.25 × 32 = 40

Content of R2 in Hex (0x) is C1200000. After converting into binary, it can be represented in IEEE- 754 format as:

1

100 0001 0

010 0000 0000 0000 0000 0000

 

Sign bit is 1 i.e. the number is negative

Biased Exponent (E’) = 100 0001 0 = 130

Normalized Mantissa (M) = 010 0000 0000 0000 0000 0000 = .25

Therefore, the number in register R1 = - 1.25 * 2^(130-127) = -1.25 * 8 = -10

R3 = R1/R2 = 40/-10 = -4

Since the number is negative, Sign bit (MSB) = 1

Converting 4 into binary of a floating point gives: (100.0)₂

Representing it into normalized form gives:  (1.000000….) × 2²

Therefore, Mantissa is 23 bits of all 0s

Biased Exponent (E’) = E+ 127 = 2+127 = 129 = (10000001)₂

It can be represented in IEEE- 754 format as:

1

100 0000 1

000 0000 0000 0000 0000 0000

 

Converting it into Hex format gives: 0x C0800000