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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 = R1/R2, what is the value stored in R3?
If R3 = R1/R2, what is the value stored in R3?
- a)0xC0800000
- b)0x40800000
- c)0x83400000
- d)0xC8500000
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider three registers R1, R2, and R3 that store numbers in IEEE-754...
R1: 0x42200000
R2: 0xC1200000
1. Subtract the exponents i.e. divisor exponent from the dividend exponent.
2. Divide the mantissa i.e.
3. Sign (opposite sign division result sign is –ve)
Hence, option (a) is correct answer.
R2: 0xC1200000
1. Subtract the exponents i.e. divisor exponent from the dividend exponent.
2. Divide the mantissa i.e.
3. Sign (opposite sign division result sign is –ve)
Hence, option (a) is correct answer.
Most Upvoted Answer
Consider three registers R1, R2, and R3 that store numbers in IEEE-754...
Solution:
Given, R1 = 0x42200000 and R2 = 0xC1200000
To find the value stored in R3 = R1/R2
Step 1: Convert R1 and R2 into decimal form
R1 = 0100 0010 0010 0000 0000 0000 0000 0000 (in binary)
= 2^8 + 2^7 + 2^2 = 256 + 128 + 4 = 388
R2 = 1100 0001 0010 0000 0000 0000 0000 0000 (in binary)
= -2^31 + 2^8 + 2^7 + 2^2 = -2147483136 + 256 + 128 + 4 = -2147482708
Step 2: Calculate R1/R2
R3 = R1/R2 = (388)/(-2147482708)
Step 3: Convert R3 into IEEE-754 single precision floating point format
Sign bit: Since R1 and R2 have opposite signs, the sign bit of R3 will be 1 (negative).
Exponent: The exponent of R3 can be calculated using the formula:
Exponent = Exponent(R1) - Exponent(R2) + Bias
where Bias = 127 (for single precision)
Exponent(R1) = 129 (in decimal) since the binary representation of R1 has an exponent of 129
Exponent(R2) = 158 (in decimal) since the binary representation of R2 has an exponent of 158
Exponent = 129 - 158 + 127 = 98 (in decimal)
The binary representation of 98 is 0110 0010. Therefore, the exponent field of R3 is 01100010.
Mantissa: To calculate the mantissa of R3, we need to divide the absolute value of R1 by the absolute value of R2.
Mantissa = 1.8091552 x 10^-10 (in decimal)
The binary representation of the mantissa is 1100 1000 1010 0001 1101 1010 1000 1111 (in binary).
Therefore, the value stored in R3 is:
1 1000010 10010001010100011101101 (in binary)
= C0800000 (in hexadecimal)
Hence, the correct answer is option A (0xC0800000).
Given, R1 = 0x42200000 and R2 = 0xC1200000
To find the value stored in R3 = R1/R2
Step 1: Convert R1 and R2 into decimal form
R1 = 0100 0010 0010 0000 0000 0000 0000 0000 (in binary)
= 2^8 + 2^7 + 2^2 = 256 + 128 + 4 = 388
R2 = 1100 0001 0010 0000 0000 0000 0000 0000 (in binary)
= -2^31 + 2^8 + 2^7 + 2^2 = -2147483136 + 256 + 128 + 4 = -2147482708
Step 2: Calculate R1/R2
R3 = R1/R2 = (388)/(-2147482708)
Step 3: Convert R3 into IEEE-754 single precision floating point format
Sign bit: Since R1 and R2 have opposite signs, the sign bit of R3 will be 1 (negative).
Exponent: The exponent of R3 can be calculated using the formula:
Exponent = Exponent(R1) - Exponent(R2) + Bias
where Bias = 127 (for single precision)
Exponent(R1) = 129 (in decimal) since the binary representation of R1 has an exponent of 129
Exponent(R2) = 158 (in decimal) since the binary representation of R2 has an exponent of 158
Exponent = 129 - 158 + 127 = 98 (in decimal)
The binary representation of 98 is 0110 0010. Therefore, the exponent field of R3 is 01100010.
Mantissa: To calculate the mantissa of R3, we need to divide the absolute value of R1 by the absolute value of R2.
Mantissa = 1.8091552 x 10^-10 (in decimal)
The binary representation of the mantissa is 1100 1000 1010 0001 1101 1010 1000 1111 (in binary).
Therefore, the value stored in R3 is:
1 1000010 10010001010100011101101 (in binary)
= C0800000 (in hexadecimal)
Hence, the correct answer is option A (0xC0800000).
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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 = R1/R2,what is the value stored in R3?a)0xC0800000b)0x40800000c)0x83400000d)0xC8500000Correct answer is option 'A'. Can you explain this answer?
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