# 程序代写代做代考 Excel mips Java javascript algorithm CSE 220: Systems Fundamentals I

CSE 220: Systems Fundamentals I

Homework #2

Spring 2017

Assignment Due: Febuary 24, 2017 by 11:59 pm

Assignment Overview

In this assignment you will be creating functions. The goal is to understand passing arguments, returning

values, and the role of register conventions. The theme of the assignment is floating point numbers and

will give you good practice manipulating register values.

You MUST implement all the functions in the assignment as defined. It is OK to implement additional

helper functions of your own in hw2.asm .

> YouMUST follow theMIPS calling and register conventions. If you do not, youWILL lose points.

> Do not submit a file with the functions/labels main or _start defined. You will obtain a ZERO for

the assignment if you do this.

� If you are having difficulties implementing these functions,write out the pseudocodeor implement the

functions in a higher-level language first. Once you understand the algorithm andwhat steps to perform,

then translate the logic toMIPS.

� When writing your program, try to comment as much as possible. Try to stay consistent with your

formatting. It is much easier for your TA and the professor to help you if we can figure out what your

code does quickly.

Getting started

Download hw2.zip fromPiazza in thehomework sectionof resources. Thisfile contains hw2.asm and

multiple hw2_main files, which you need for the assignment. At the top of your hw2.asm program in

comments put your name and SBU ID number.

# Homework #2

# name: MY_NAME

# sbuid: MY_SBU_ID

CSE 220 – Spring 2017 Homework #2 Page 1

How to test your functions

To test your functions, simply open one of the provided hw2_main files in MARS. Next, assemble the

main file and run. Mars will take the contents of the file referenced with the .include at the end

of the file and add the contents of your hw2.asm file to the main file before assembling it. Once the

contents have been substituted into the file, Mars will then assemble it as normal.

Each of the main files tests the functions you are to implement with one of the sample test cases. You

shouldmodify these files or create your own files, to test your functions withmore test cases.

> Your assignment will not be graded using these tests!

Any modifications to the main files will not be graded. You will only submit your hw2.asm file via

Sparky. Make sure that all code require for implementing your functions ( .text and .data ) are in-

cluded in the hw2.asm file! Tomake sure that your code is self-contained, try assemblingyour hw2.asm

file by itself in MARS. If you get any errors (such as a missing label), this means that you need to refactor

(reorganize) your code, possibly bymoving labels you inadvertently defined in a main file to hw2.asm .

> It is highly advised to write your own main programs (new individual files) to test each of your func-

tions thoroughly.

> Make sure to initialize all of your valueswithin your functions! Never assume registers ormemorywill

hold any particular values!

Optional readings

If you are interested there are a few interesting readings about floating point arithmetic which contain

information that youmay not be aware of and can be used to assist youwith the assignment.

• What every computer programmer should know about floating point

• What Every Computer Scientist Should KnowAbout Floating-Point Arithmetic

• Floating-point arithmetic may give inaccurate results in Excel

• What Every JavaScript Developer Should KnowAbout Floating Point Numbers

Part I: Helper Functions

In this part of the assignment you will implement basic leaf functions (a function which does not call any

other functions). hw2_main1.asm contains a basic test for each of these functions.

All functions implemented in the assignment must be placed in hw2.asm and follow the standardMIPS

register conventions for functions taught in lecture.

CSE 220 – Spring 2017 Homework #2 Page 2

http://blog.reverberate.org/2014/09/what-every-computer-programmer-should.html?m=1

http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

http://support.microsoft.com/kb/78113

http://blog.chewxy.com/2014/02/24/what-every-javascript-developer-should-know-about-floating-point-numbers/

a. int char2digit(char c)

This function converts the ASCII character ’0’-’9’ to its integer value, 0-9. Any other ASCII character

results in an error.

• c : the ASCII character

• returns: the integer value of the character or -1 for error

Examples:

Code Return Value

char2digit(‘9’) 9

char2digit(‘3’) 3

char2digit(‘@’) -1

char2digit(‘c’) -1

b. int memchar2digit(char *c)

This function converts the ASCII character stored at the memory address given by c from ’0’-’9’ to

its integer value, 0-9. Any other ASCII character results in an error.

� The * is C notation for an address of a value stored inmemory.

• *c : the address of an ASCII character in memory

• returns: the integer value of the character or -1 for error

Examples:

Note, the prefix 0x indicates the following digits are shown in hexadecimal representation.

Code Memory Address = Stored Value Return Value

memchar2digit(0x40000000) Mem[0x40000000] = 0x39 9

memchar2digit(0x44404440) Mem[0x44404440] = 0x33 3

memchar2digit(0x4FFFFFFF) Mem[0x4FFFFFFF] = 0x4B -1

memchar2digit(0x30303030) Mem[0x30303030] = 0x20 -1

c. (int, int) fromExcessk(int value, int k)

This function converts an excess-k value to its corresponding value in base-10.

• value : an excess-k value. If value is≤ 0, an error occurs.

• k : a integer value representing excess-k. If k ≤ 0, an error occurs.

• returns: (i) 0 for success, -1 for error, and (ii) the converted base 10 value upon success, the

original value upon error. Note that the function returns two values: $v0 holds the first

return value and $v1 holds the second return value.

CSE 220 – Spring 2017 Homework #2 Page 3

> Assume all input and resultant values are correctly representable in 32-bits.

Examples:

Code Return Value

fromExcessk(23, 127) 0, -104

fromExcessk(-23, 127) -1, -23

fromExcessk(10, 35) 0, -25

fromExcessk(1000,-10) -1, 1000

fromExcessk(2,0) -1, 2

d. int printNbitBinary(int value, int m)

This function prints out the m least-significant bits of the binary representation of value and re-

turns success. If m is in the range [1, 32] the function prints out the m least-significant bits of the

binary representation of the value and returns success (0). Otherwise, the function prints nothing

and returns error (-1).

• value : the value to print the binary digits of

• m : the number of bits to print of value .

• returns: 0 if the print was successful, -1 for error.

> Remember, value is stored in 2’s complement binary format in the register.

> value is any 32-bit value in the register.

Basic Algorithm:

shift value left (32-m) bits

while (m > 0)

if (value < 0) print ‘1’, else print ‘0’ shift value left 1 bit m = m-1 Examples: Code Return Value Prints printNbitBinary(-23, 30) 0 111111111111111111111111101001 printNbitBinary(10, 3) 0 010 printNbitBinary(10, 9) 0 000001010 printNbitBinary(1000,-2) -1 printNbitBinary(2,100) -1 CSE 220 – Spring 2017 Homework #2 Page 4 Part II: Floating point In this section, you will be writing functions to work with strings and floating point values and for- mats. Youwill implement the functions to complete hw2_main2.asm . These functionswill call the functions you implemented in Part I. All functions implemented in the assignment must be placed in hw2.asm and follow the standard MIPS register conventions for functions taught in lecture. < Under no circumstances are you to use the floating point registers or floating point MIPS instructions. e. (int, float) btof(char[] input) � The [] is C notation for the starting address of an array of the specified data type stored inmemory. btof will convert a string of binary digits or a string of characters representing a special value into its IEEE 754 single precision floating point binary representation. The function will return two values, an int and a ‘float‘. The int is a flag for success or error. The “float" is the IEEE-754 single floating point representation of the input value stored in the returned register. • input : a string of binary digits or special value string representing an IEEE-754 number. Can be of any length. The IEEE representation should be truncated to fit in IEEE-754 single preci- sion format. • returns: (i) 0 conversionwas successful, -1 for error and (ii) the IEEE-754 formatted value of the string, or unspecified value if an error occurred. � Your implementationmay return any value youwant for the unspecified value. This functionmust parse the string until it reaches the end (‘’) or an invalid character. Valid binary digit characters are 0 , 1 , . , + , - , and the ASCII characters in NaN and Inf . If any non-valid character/substring is reached, an error occurs. TheSPECIALfloat valueswhichmustbehandledare +0.0 , -0.0 , NaN , +Inf , and -Inf . These strings are CASE-SENSITIVE. < Your functionMUSTCALL char2digit or memchar2digit . < All input argument strings, which are not SPECIAL float values, will always contain a single radix point (‘.‘) and have at least 1 binary digit on either side of the radix point. Examples: Note, the prefix 0b indicates the following digits are the binary representation for the floating point representation. This notation is similar to the hexadecimal representation 0x . CSE 220 – Spring 2017 Homework #2 Page 5 Code Return Values btof("010.001") 0, 0b01000000000010000000000000000000 btof("+010.001") 0, 0b01000000000010000000000000000000 btof("-010.001") 0, 0b11000000000010000000000000000000 btof("-0.0001") 0, 0b10111101100000000000000000000000 btof("11110.001110001110001110001") 0, 0b01000001111100011100011100011100 btof("-11110.001110001110001110001") 0, 0b11000001111100011100011100011100 btof("+0.0") 0, 0b00000000000000000000000000000000 btof("-0.0") 0, 0b10000000000000000000000000000000 btof("NaN") 0, 0b01111111111111111111111111111111 btof("+Inf") 0, 0b01111111100000000000000000000000 btof("-Inf") 0, 0b11111111100000000000000000000000 btof("213.2") -1, ? btof("iNF") -1, ? btof("nan-12.3") -1, ? btof("-11.1A") -1, ? btof("@.2") -1, ? � ? denotes an unspecified return value. Your implementation may return any value you want. f. int print_parts(float value) This function takes the binary representation of a single precision IEEE-754 float value as a param- eter and returns a 1 if the value is positive, 0 if the value is a special value ( +0 , -0 , NaN , +Inf , and -Inf ), and -1 if the value is negative. This function will print out the sign of the number, the binary and decimal value of the exponent, and the binary and decimal value of themantissa. < Your functionMUSTCALL printNbitBinary . < Your functionMUST print the EXACT format shown in the examples. All spaces are single spaces. < DONOT use system call 35 to print the binary. It always prints out 32 bits. • value : the binary representation of a floating point value. • returns: 1 if the value is positive, 0 if the value is a special value, and -1 if the value is negative Examples - Note the argument is shown as a hexadecimal value to save space: CSE 220 – Spring 2017 Homework #2 Page 6 Code Return Value Prints print_parts(0x40080000) 1 0 + Value: 2.125 10000000 128 00010000000000000000000 524288 print_parts(0xC1F1C71C) -1 1 - Value: -30.222222 10000011 131 11100011100011100011100 7456540 print_parts(0x7F800000) 0 0 + Value: +Inf 11111111 255 00000000000000000000000 0 � Use the IEEE 754 Converter to assist with checking your values. g. int print_binary_product(float value) This function takes the binary representation of a single precision IEEE-754float value as a parame- ter prints the value in a binary product format (which is similar to scientific notation):±1.mantissa ∗ 2+/−exponent if it is a non-special value. The function will return 1 if the value is printable in binary product form or 0 if the value is a special value. • value : the binary representation of a floating point value. • returns: 1 if the value is printable in binary product form or 0 if the value is a special value < Your functionMUSTCALL printNbitBinary and fromExcessk . < Your functionMUST print the EXACT format shown in the examples. All spaces are single spaces. Examples - Note the argument is shown as a hexadecimal value to save space: Code Rtn Val Prints print_binary_product(0x40080000) 1 +1.00010000000000000000000 x 2^+1 print_binary_product(0xC1F1C71C) 1 -1.11100011100011100011100 x 2^+4 print_binary_product(0xBDDC0000) 1 -1.10111000000000000000000 x 2^-4 print_binary_product(0x3DDC0000) 1 +1.10111000000000000000000 x 2^-4 print_binary_product(0x7F800000) 0 Hand-in Instructions See Sparky Submission Instructions on piazza for hand-in instructions. < There is no tolerance for homework submission via email. Work must be submitted through Sparky. Please do not wait until the last minute to submit your homework. If you are struggling, stop by office hours for additional help. CSE 220 – Spring 2017 Homework #2 Page 7 http://www.binaryconvert.com/result_float.html