--- tags: computer-arch --- # Quiz1 of Computer Architecture (2021 Fall) > Solutions :::info :information_source: General Information * You are allowed to read [lecture materials](http://wiki.csie.ncku.edu.tw/arch/schedule). * That is, an open book exam. * You shall not disclose your answer during the quiz. * Each answer has 5 points. * :timer_clock: 09:10 ~ 10:20AM on Oct 5, 2021 ::: ## Problem A [Two's Complement](https://en.wikipedia.org/wiki/Two%27s_complement) is the standard for representing signed integers: * The most significant bit (MSB) has a negative value; all others have positive values (same as unsigned) * Binary addition is performed the same way for signed and unsigned * The bit representation for the negative (additive inverse) of a two's complement number can be found by: - flipping all the bits and adding 1 (i.e.-x = ~x + 1). ![](https://hackmd.io/_uploads/ry-ws0uEt.png) The above "number wheel" showing the relationship between 4-bit numerals and their [Two's Complement](https://en.wikipedia.org/wiki/Two%27s_complement) interpretations is shown on the right: * The largest number is 7 whereas the smallest number is -8 * There is a nice symmetry between numbers and their negative counterparts except for -8 This section is designed as a conceptual check for you to determine if you conceptually understand and have any misconceptions about this topic. Please answer **Yes** / **No** to the following questions: 1. Is it possible to get an overflow error in [Two's Complement](https://en.wikipedia.org/wiki/Two%27s_complement) when adding numbers of opposite signs? > * A01 = ? ==No==. > Overflow errors only occur when the correct result of the addition falls outside the range of $[−(2^{n−1}), 2^{n−1} − 1]$. Adding numbers of opposite signs will not result in numbers outside of this range. 2. If you interpret a $N$ bit [Two's Complement](https://en.wikipedia.org/wiki/Two%27s_complement) number as an unsigned number, would negative numbers be smaller than positive numbers? > * A02 = ? ==No== > In Two's Complement, the MSB is always 1 for a negative number. This means ALL Two's Complement negative numbers will be larger than the positive numbers. 3. If you interpret an $N$ bit [Bias notation](https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-01-unified-engineering-i-ii-iii-iv-fall-2005-spring-2006/comps-programming/number_systems.pdf) number as an unsigned number (assume there are negative numbers for the given bias), would negative numbers be smaller? > * A03 = ? ==Yes== > In bias notation, we add a bias to the unsigned interpretation to create the value. This means that negative numbers will stay smaller than positive numbers. This is unlike Two's Complement. --- ## Problem B Arithmetic overflow occurs when the result of a calculation can't be represented in the current encoding scheme (i.e., it lies outside of the representable range of values), resulting in an incorrect value. * Unsigned overflow: the result lies outside of $[UMin, UMax]$; an indicator of this is when you add two numbers and the result is smaller than either number. * Signed overflow: the result lies outside of $[TMin, TMax]$; an indicator of this is when you add two numbers with the same sign and the result has the opposite sign. ![](https://hackmd.io/_uploads/S1ndpR_EY.png) 1. Find the largest 8-bit unsigned numeral c (answer in HEX) such that c + 0x80 causes NEITHER signed nor unsigned overflow in 8 bits. > * B01 = ? ==0x7F== > Unsigned overflow will occur for c > 0x80. Signed overflow can only happen if c is negative (also > 0x80). Largest is therefore, 0x7F 2. Find the smallest 8-bit numeral c (answer in HEX) such that c + 0x71 causes signed overflow, but NOT unsigned overflow in 8 bits. > * B02 = ? ==0xF== > For signed overflow, need (+) + (+) = (−). For no unsigned overflow, need no carryout from MSB. The first (−) encoding we can reach from 0x71 is 0x80. 0x80 – 0x71 = 0xF. --- ## Problem C According to IEEE 754 Floating Point Standard, the value of a real number can be represented in scientific binary notation as: $$Value = (-1)^{sign} \times Mantissa_{2} \times 2^{Exponent} = (-1)^S \times 1.M_2 \times 2^{E-bias}$$ The binary representation for floating point values uses three fields: * S: encodes the sign of the number (0 for positive, 1 for negative) * E: encodes the exponent in **biased notation** with a bias of $2^{w-1}-1$ * M: encodes the mantissa (or significand, or fraction) – stores the fractional portion, but does not include the implicit leading 1. ![](https://hackmd.io/_uploads/HkgOlyKNK.png) 1. Let’s say that we want to represent the number 3145728.125~10~ (broken down as $2^{21} + 2^{20} + 2^{−3}$). Is it enough to represent this number single precision floating point? (Please answer **Yes** / **No**) > * C01 = ? ==No== > Could only represent $2^{21} + 2^{20}$. Not enough bits in the mantissa to hold $2^{-3}$, which caused rounding. 2. What is the decimal value of float 0x80000000? (Answer with leading + and -) > * C02 = ? ==-0== 3. What is the decimal value of float 0xFF94BEEF? (Answer with leading + and -) > * C03 = ? ==NaN== 4. What is the decimal value of float 0x41180000? (Answer with leading + and -) > * C04 = ? ==+9.5== > 0x41180000 = 0b 0|100 0001 0|001 1000 0…0. > S = 0, E = $128+2 = 130$ $\to$ Exponent = E – bias = 3, Mantissa = 1.0011~2~ > $$> 1.0011_2 \times 2^3 = 1001.1_2 = 8 + 1 + 0.5 = 9.5 >$$ 5. What is the smallest positive value that can be stored using a single precision float? Answer in HEX value. (video: [How to Calculate Smallest Float Value in IEEE 754 Standard (Single Precision)](https://youtu.be/SMLcrWgE2sk)) > * C05 = ? ==0x00000001== > 0x00000001 = $2^{−23} \times 2^{−126}$ 6. What is the smallest positive normalized value that can be stored using a single precision float? > * C06 = ? ==0x00800000== > 0x00800000 = $2^{−126}$ 7. What is the decimal value of float 0xFF800000? (Answer with leading + and -) > * C07 = ? ==-∞== 8. What is the decimal value of float 0x421E4000? (Answer with leading + and -) > * C08 = ? ==+39.5625== --- ## Problem D Floating Point Mathematical Properties * Not associative: $(2 + 2^{50}) – 2^{50} \neq 2 + (2^{50} – 2^{50})$ * Not distributive: $100 \times (0.1 + 0.2) \neq 100 \times 0.1 + 100 \times 0.2$ * Not cumulative: $2^{25} + 1 + 1 + 1 + 1 \neq 2^{25} + 4$ If x and y are variable type float, will the expression (x+2*y)-y==x+y always be evaluated as true? (Please answer **Yes** / **No** with explanation.) > D01 = ? ==No, Rounding error / Overflow== (後者只要一個解釋符合就給分) --- ## Problem E Compute the decimal result of the following arithmetic expressions involving **6-bit** [Two's Complement](https://en.wikipedia.org/wiki/Two%27s_complement) numbers as they would be calculated on a computer. Do any of these result in an overflow? Are all these operations possible? 1. Input: 0b100011 + 0b111010 > * E01 = ? ==Overflow== > Adding together we get 0b1011101, however since we are working with 6-bit numbers we truncate the first digit to get 0b011101 = 29. Since we added two negative numbers and ended up with a positive number, this results in an overflow. 2. Input: 0xFF − 0xAA > * E02 = ? ==Impossible== > This is not possible, as these hex numbers would need 8 bits to represent and we are working with 6 bit numbers. 3. Input: 0x3B + 0x06 > * E03 = ? ==1== > Converting to binary, we get 0b111011 + 0b000110 = (after truncating) 0b000001 = 1. > Despite the extra truncated bit, this is not an overflow as -5 + 6 indeed equals 1 --- ## Problem F 1. How do you write the bitwise [exclusive-nor (XNOR)](https://en.wikipedia.org/wiki/XNOR_gate) operator in C program? > * F01 = ? ==x == y== (只要有比較就給分) 2. Given x as an unsigned integer. Please use bitwise operators and +, -, == to check if x is a power of 2. Write down the expression in C without any branches (i.e., if, else, ? :, do, while, for, goto) > * F02 = ? ==x & (x - 1) == 0== 3. The following function absf returns absolute value of a single precision float. What is the value of F03? Answer in HEX. cpp #include <stdint.h> float absf(float x) { uint32_t mask = F03; union { uint32_t i; float f; } u = {.f = x}; u.i &= mask; return u.f; }  > * F03 = ? ==0x7fffffff==