CS代考 CITS2002 Systems Programming – cscodehelp代写

SEMESTER 2, 2020 EXAMINATIONS Physics, Mathematics & Computing
Department of Computer Science and Software Engineering
This paper contains: 6 Pages (including title page)
CITS2002 Systems Programming
Programming and Systems Time Allowed: 2:00 hours
This Paper Contains 6 Pages and 4 Questions.
You are required to attempt ALL FOUR (4) questions.
Time Allowed: 2 hours.
You may use any number of blank pages for rough working. You do not need to submit these pages at the end of the exam.
No other materials, such as textbooks, notes, or calculators, are permitted during the exam.
SUPPLIED STATIONERY ALLOWABLE ITEMS 1 x Answer Booklet 18 Pages No Allowable Items.
Examination candidates may only bring authorised materials into the examination room. If a supervisor finds, during the examination, that you have unauthorised material, in whatever form, in the vicinity of your desk or on your person, whether in the examination room or the toilets or en route to/from the toilets, the matter will be reported to the head of school and disciplinary action will normally be taken against you. This action may result in your being deprived of any credit for this examination or even, in some cases, for the whole unit. This will apply regardless of whether the material has been used at the time it is found. Therefore, any candidate who has brought any unauthorised material whatsoever into the examination room should declare it
to the supervisor immediately. Candidates who are uncertain whether any material is authorised should ask the supervisor for clarification.
Candidates must comply with the Examination Rules of the University and with the directions of supervisors. No electronic devices are permitted during the examination.
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1) In the C99 programming language, a colour can be represented by 3 integer values, each between 0 and 255, storing the magnitude of the colour’s red, green, and blue components.
One measure of the distance between any two colours, is the sum of the squares of the differences between their red, green, and blue components. For example, if the Colour1 is represented by the integers R1, G1, and B1, and Colour2 is represented by the integers R2, G2, and B2, then the distance between the two colours may be expressed as:
distance = (R2-R1)*(R2-R1) + (G2-G1)*(G2-G1) + (B2-B1)*(B2-B1);
The smaller the distance, the closer are the two colours. Consider the C99 function whose prototype is:
char *closestRGB(char *filename, int R1, int G1, int B1);
The function receives 4 parameters − the name of a text file containing many colours (one per
line), and three integers. The text file might typically contain:
190 190 190 grey
65 105 225 RoyalBlue
0 255 255 cyan 0100 0 DarkGreen
where each line provides the red, green, and blue components of the named colour. Each line’s fields are separated by whitespace characters.
The purpose of the function is to return a pointer to dynamically allocated memory storing the name of the colour closest to the colour represented by the function’s integer parameters. If the function detects any problems, the function should simply return NULL.
Write the closestRGB() function in C99.
2) One data-structure that has grown in prominence in newer programming languages is the hashtable (or hashmap, or dictionary). They are similar to arrays, in that we use an index to insert or find a required item. However, unlike the standard arrays in C99 where the index is an integer, a hashtable’s index may be of another datatype, such as a string.
Unlike some other programming languages, C99 does not provide a hashtable data-structure, but one can be implemented using a few simple functions that manage dynamically allocated memory.
Assuming that you have a function that hashes a string to an integer, with the prototype:
unsigned int hash_string(char *string);
develop a simple hashtable data-structure in C99.
Firstly, using C99’s typedef feature, define a user-defined type named HASHTABLE to represent your data-structure. Your data type should employ a fixed sized array of pointers, each of which points to a list.
Each list consists of linked structures, with each structure storing a value (here, a string), and a pointer to a structure of the same datatype. Each linked list is terminated by a NULL- pointer.
Your data-structure should be managed and accessed by two functions, with the following C99 prototypes:
bool add_to_hashtable(HASHTABLE *hashtable, char *string); and
bool find_in_hashtable(HASHTABLE *hashtable, char *string); On success each function returns true, or returns false if the function detects any problem
with its parameters or during its execution.
Write add_to_hashtable() and find_in_hashtable() in C99.
3) make is a system utility commonly used to build programming projects. make first determines the sequence of commands to be executed to build a project. The commands are then executed, in turn, until all commands have executed successfully, or until one of them is unsuccessful.
make has no knowledge of the commands (and their arguments) that require executing. Instead, make invokes the standard shell to execute each command. For example, if make determines that an action involving the compilation of a C99 program must be performed, then make may execute a command such as:
/bin/sh −c “cc −o testing.o testing.c”
by passing the required command (action) as a single string to the shell, and then capturing the exit status of the shell to determine if the command was successful, or not.
Consider the C99 function whose prototype is:
int executeCommandSequence( char *commands[] );
The function receives a vector of character strings, each of which is passed to a new shell
process to be executed. The vector of strings is terminated by a NULL pointer.
The commands are executed, in turn, until all commands have executed successfully (in which case the function returns 0), or until one of them is unsuccessful (in which case the function returns the exit status of the unsuccessful command).
Write the executeCommandSequence() function in C99.
4a) Explain what information must be managed within a Unix-based operating system kernel when a process invokes a fork() system-call followed shortly thereafter by a wait() system-call, and the new child process invokes an execve() system-call and eventually invokes an exit() system-call.
4b) When computer systems first supported multi-programming, the memory required for each process was initially allocated using equal-sized memory partitioning. While successful, this approach exhibited some problems that were later overcome with the introduction of unequal- sized memory partitioning.
i) Briefly describe two problems resulting from the use of fixed-sized partitioning, and describe how these were addressed by using unequal-sized partitions.
After several years of experimentation with different memory partitioning approaches, contemporary computer systems now almost exclusively employ a fixed-sized memory partitioning scheme termed paging.
ii) Briefly explain how paging is able to overcome the disadvantages of both traditional fixed- sized partitioning and unequal-sized partitioning.

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