Computer Arithmetic Questions and Answers

Answer: A

The flags are S = 1, Z = 0, AC = 0, P = 0, Cy = 0.

\[ \begin{aligned} 47\text{H} &= 0100\,0111 \\ 51\text{H} &= 0101\,0001 \\ \text{sum} &= 1001\,1000 = 98\text{H} \end{aligned} \]

• S = 1: bit 7 of the result is 1.
• Z = 0: the result is not zero.
• AC = 0: the low nibbles 7 + 1 = 8 produce no carry out of bit 3.
• P = 0: 98H contains three 1s, an odd number, and the 8085 sets P only for even parity.
• Cy = 0: 98H fits in 8 bits, so there is no carry out of bit 7.

Options B and C set Z = 1, which is impossible for a non-zero result, so neither they nor 'None of the above' fit.

Answer: D

The result is \(-\ \text{579}_{H}\).

In 15's complement subtraction, add the 15's complement of the subtrahend. The 15's complement of C14 is found by subtracting each digit from F: F - C = 3, F - 1 = E, F - 4 = B, giving 3EB.

\[ \begin{aligned} \text{69B} + \text{3EB} &= \text{A86} \\ \text{end carry} &= \text{none} \end{aligned} \]

With no end-around carry, the result is negative and in complement form. Its magnitude is the 15's complement of A86: F - A = 5, F - 8 = 7, F - 6 = 9, so the answer is -579 hexadecimal. Check: C14 - 69B = 579.

3EB is only the complement of C14, A86 is the uncorrected sum, and 579 without the sign is the magnitude, so all three miss the sign or the final step.

Answer: C

CB2H - 972H = 340H.

In 16's complement subtraction, add the 16's complement of the subtrahend and discard the end carry. The 15's complement of 972 is FFF - 972 = 68D; adding 1 gives the 16's complement 68E.

\[ \begin{aligned} \text{CB2} + \text{68E} &= \text{1340} \\ \text{discard carry} &\Rightarrow \text{340H} \end{aligned} \]

Check in decimal: CB2H = 3250, 972H = 2418, and 3250 - 2418 = 832 = 340H. 68DH is only the 15's complement of the subtrahend, and 1340H is the sum before the end carry is dropped.

Answer: B

The exponent bias in IEEE-754 single precision is 127.

A single-precision number has 1 sign bit, an 8-bit exponent and a 23-bit fraction. The exponent is stored with a bias so that it is always non-negative, and the bias is

\[ 2^{8-1} - 1 = 127 \]

so a true exponent of 0 is stored as 127. Double precision has an 11-bit exponent, giving a bias of 2¹⁰ - 1 = 1023, which is the trap in option C. 128 and 1024 are off by one: the bias is one less than a power of two, which leaves the all-zeros and all-ones exponent patterns for zero, subnormals, infinity and NaN.

Answer: C

0.25 in binary is 0.01.

Convert a decimal fraction by repeated multiplication by 2, taking the integer part each time:

\[ \begin{aligned} 0.25 \times 2 &= 0.5 \Rightarrow 0 \\ 0.5 \times 2 &= 1.0 \Rightarrow 1 \end{aligned} \]

The fraction reaches 0, so the digits read downward give 0.01. Check: the second binary place is worth 1/4, and 1/4 = 0.25. The other options are other quarters and eighths: 0.10 is 1/2 = 0.5, 0.11 is 3/4 = 0.75 and 0.001 is 1/8 = 0.125.

Answer: A

Single precision allocates 8 bits to the exponent.

An IEEE 754 single-precision number is 32 bits: 1 sign bit, an 8-bit exponent stored with a bias of 127, and a 23-bit fraction (mantissa), with an implied leading 1 for normal numbers.

23 bits is the fraction field, not the exponent. 11 bits and 52 bits are the exponent and fraction of double precision (64 bits: 1 + 11 + 52).

Answer: A

The 8-bit two's complement of -18 is 11101110.

Write +18 in 8 bits, invert every bit (one's complement), then add 1.

\[ \begin{aligned} +18 &= 00010010 \\ \text{invert} &= 11101101 \\ +1 &= 11101110 \end{aligned} \]

Check: 11101110 is 238 as an unsigned number, and 238 - 256 = -18. 11101101 is the one's complement, which stops before adding 1; 11101111 and 11101100 are one off in the other direction, and 11101010 is -22.

Answer: B

The 2's complement of 0110 is 1010.

Invert every bit to get the one's complement, then add 1.

\[ \begin{aligned} 0110 &\rightarrow 1001 \\ 1001 + 1 &= 1010 \end{aligned} \]

Check: 0110 is +6, and in 4-bit two's complement 1010 is -8 + 2 = -6. 1001 is only the one's complement, 1100 is -4, and 0110 is the number unchanged.

Answer: B

The signed binary number 1101010 in 2's complement form has a leading 1, so it is negative. To find its magnitude, invert the bits to get 0010101 and add 1 to get 0010110, which is 16 + 4 + 2 = 22. So the value is -22, option (B). The value -42 in (A) is what is found by reading the number as a sign and magnitude, 1 followed by 101010 = 42, and -106 in (D) comes from taking the unsigned 1101010 = 106 and adding a negative sign. The value -21 in (C) results from inverting the bits without adding 1.