program程序讲解、讲解C++程序语言

日期：2024-02-25 08:05

Assignment 1
Introduction
In Assignment 1, students will learn to use system calls.
System Calls
Manage Files
int open(const char *pathname, int flags, ...
/* mode_t mode */ );
ssize_t read(int fd, void buf[.count], size_t count);
ssize_t write(int fd, const void buf[.count], size_t count);
open()
The open() system call opens the file specified by pathname. If the specified file does not exist,
it may optionally (if O_CREAT is specified in flags) be created by open().
The return value of open() is a file descriptor, a small, nonnegative integer that is an index to an
entry in the process's table of open file descriptors. The file descriptor is used in subsequent
system calls (read(2), write(2), lseek(2), fcntl(2), etc.) to refer to the open file. The file descriptor
returned by a successful call will be the lowest-numbered file descriptor not currently open for
the process.
read()
read() attempts to read up to count bytes from file descriptor fd into the buffer starting at buf.
On files that support seeking, the read operation commences at the file offset, and the file offset
is incremented by the number of bytes read. If the file offset is at or past the end of file, no bytes
are read, and read() returns zero.
If count is zero, read() may detect the errors described below. In the absence of any errors, or if
read() does not check for errors, a read() with a count of 0 returns zero and has no other effects.
write()
write() writes up to count bytes from the buffer starting at buf to the file referred to by the file
descriptor fd.
The number of bytes written may be less than count if, for example, there is insufficient space on
the underlying physical medium, or the RLIMIT_FSIZE resource limit is encountered
(seesetrlimit(2)), or the call was interrupted by a signal handler after having written less than
count bytes. (See also pipe(7).)
For a seekable file (i.e., one to which lseek(2) may be applied, for example, a regular file) writing
takes place at the file offset, and the file offset is incremented by the number of bytes actually
written. If the file was open(2)ed with O_APPEND, the file offset is first set to the end of the file
before writing.
The adjustment of the file offset and the write operation are performed as an atomic step.
Map files
mmap() creates a new mapping in the virtual address space of the calling process. The starting
address for the new mapping is specified in addr. The length argument specifies the length of
the mapping (which must be greater than 0).
For more details, see mmap(2) - Linux manual page
void *mmap(void addr[.length], size_t length, int prot, int flags,int fd, off_t
offset);
int munmap(void addr[.length], size_t length);
Process
pid_t fork(void);
pid_t wait(int *_Nullable wstatus);
pid_t waitpid(pid_t pid, int *_Nullable wstatus, int options);
int waitid(idtype_t idtype, id_t id, siginfo_t *infop, int options);
fork()
fork() creates a new process by duplicating the calling process.
The new process is referred to as the child process. The calling process is referred to as the
parent process.
The child process and the parent process run in separate memory spaces. At the time of fork()
both memory spaces have the same content.
Memory writes, file mappings (mmap(2)), and unmappings
(munmap(2)) performed by one of the processes do not affect the other.
The child process is an exact duplicate of the parent process except for the following points:
? The child has its own unique process ID, and this PID does not match the ID of any existing
process group (setpgid(2)) or session.
? The child's parent process ID is the same as the parent's process ID.
? The child does not inherit its parent's memory locks (mlock(2), mlockall(2)).
For more details, see fork(2) - Linux manual page
wait()
All of these system calls are used to wait for state changes in a child of the calling process, and
obtain information about the child whose state has changed. A state change is considered to be:
the child terminated; the child was stopped by a signal; or the child was resumed by a signal. In
the case of a terminated child, performing a wait allows the system to release the resources
associated with the child; if a wait is not performed, then the terminated child remains in a
"zombie" state (see NOTES below).
If a child has already changed state, then these calls return immediately. Otherwise, they block
until either a child changes state or a signal handler interrupts the call (assuming that system
calls are not automatically restarted using the SA_RESTART flag of sigaction(2)). In the
remainder of this page, a child whose state has changed and which has not yet been waited
upon by one of these system calls is termed waitable.
Graph Preleminary
Graph Layout
OA : Offset Array.
NA : Neighbor Array.
In our implementation, the graph is in CSR format. Which means that, if you want to know the
neighbor of node i , get OA[i] and OA[i+1] , the index range of NA that stores
neighbors.
In the graph.size file, it has 8 bytes, which are the number of nodes and number of edges.
( |nodes| , |edges| )
In the graph file, the first |nodes|+1 *4 Bytes are values of OA . The following |edges| *4
Bytes are values of NA . The last |edges| *4 Bytes are values of edge weight for
corresponding edge.
Dataset Download
You can get dataset, and sample output of Assignment 1 from the following link:
CSC3150_23-24Term2_A1_data - OneDrive
Your dataset should be placed at the same directory of code
main@ubuntu:~/Desktop$ ls
a.out grader g1 g1.size g2 g2.size graph.cpp graph.h sample_output
Important
In this assignment, all datasets to read and output files you need to write are in binary format.
Please take a look at std::ios::binary .
sample_output is used for self-checking if the answers to task1 are correct.
Task1
In task1, you are provided with data sets graph1 , graph2 correspondings to Task1_1
and Task1_2 . In such graphs, the nodes can be represented as cities, edges between cities
can be represented as roads. We provide a program traversing nodes through BFS algorithm.
Figure: Transportation Graph
? Your first task is to complete the remaining part of program so that it can correctly load data
into memory and do BFS traverse execution.
? After BFS traversal is done, the program needs to store the sequence of visits to the result file.
? In the following progress charts, blocks in yellow means it's a file, in red means it's a TODO
part for you.
? In this task, you need to complete Task1_1(), Task1_2() . In Task1_2() we only
have 2GB memory but g2 has 1.8GB data. Traditional read from disk to memory does not
work.
Hint
? When loading the graph, please note that OA has length |nodes|+1 instead of
|nodes| . This is because we miantain one more integer for some boundary problems.
What you read from the .size file is |nodes| and |edges| .
Task2
In task 2, the program goes into a simple simulation. In the daytime, a node gains weights from
its edges with corresponding edge weights, implemented by produce() . In the night, a
node's weight will self-increased by 1, implemented by consume() .
The global view of process is shown in the figure below. We highlight the components you needs
to do in green and yellow blocks:
You are required to use fork() to handle two processes. For the parent process, we will do
calculation of produce() . For the child process, we will do consume() . Both of these
functions are provided, you do not need to do any modification to it.
// Executed in parent process
void produce(int *weights, int len, Graph &g) {
for (int i=0; i<len; i++) {
int l,r;
g.get_edge_index(i,l,r);
for (int j=l;j<r;j++) {
weights[i+1]+=g.edge_weights[j];
}
}
}
// Executed in child process
void consume(int *weights, int len, Graph &g) {
for (int i=0; i<len; i++) {
weights[i+1] += 1;
}
}
TODO
? Your task is to complete the following functions with correct system call and control logic so
that program runs correctly as shown in the above figrue.
? In implementation of communication, it actually shares the data stored in weights
? Format of array weights : the first integer should be the value of current iteration (start
from 0). The following |V| integers are corresponding weights of vertex
? For each iteration generated by child process, child process should write result to file
e.g. For 4th and 5th iterations, write sequence of nodes weight to
/home/csc3150/A1/Task2_parent.output_2 and
/home/csc3150/A1/Task2_child.output_2 .
? Program terminates at 10th iteration
// Task2: Process Inter-Communicationvoid Task2() {
Graph g;
int fd;
fd = g.map_global_graph("g2");
std::string ipc_path("ipc_file");
// Creating inter process communication file if there is not.
int ipc_fd = open(ipc_path.c_str(),O_RDWR| O_CREAT,0777);
lseek (ipc_fd, (g.v_cnt+1)*sizeof(int)-1, SEEK_SET);
write (ipc_fd, "", 1);
close(ipc_fd);
std::string output_parent_path("Task2_parent.output");
std::string output_child_path("Task2_child.output");
// process control
}
// Task2: Process Inter-Communication with control
void parent_process(const std::string &path) {
int pid = getpid();
printf("parent proc %d is producing\n",pid);
// produce()
return;
}
// Task2: Process Inter-Communication with control
void child_process(const std::string &path) {
int pid = getpid();
printf("child proc %d is consuming\n",pid);
// consume()
return;
}
Hint
? The status of vertices need to be communicated inter-process at each iteration because both
processes need to process/ fetch data.
? You might need to define index 0 of mapped file in mmap() as control int
? Because of Data Dependency, execution order matters
? By reading control byte, each process will know whether they should wait, process, or
terminate.
? You can use mmap() to share reading the data graph if both processes need information of
graph.
? Try to figure out how memory is consumed when multi-process using mmap map to one
file.
Output files
You are expected to generate the following files when you complete all the tasks
All these files are in the same directory as graph.h, graph.c
(wrong file name will not be correctly graded):
Report 10'
You shall strictly follow the provided latex template for the report, where we have emphasized
important parts and respective grading details. Reports based on other templates will not be
graded.
Report Template
? You can find latex template in the following link
? https://www.overleaf.com/read/ybgnjwnyvjpx#dd17ed
? We also provide template at BB.
LaTex Editor
For your convenience, you might use Overleaf, an online LaTex Editor.
1. Create a new blank project.
2. Click the following highlight bottom and upload the template we provide.
3. Click Recompile and you will see your report in PDF format.
Extra Credit
Instead of using shared memory space for inter-process communication, you are required to
implement Task2 with using message passing.
Using any profiling tools to compare performance of communication via shared memory
(mmap) and message passing.
Report your result and your discovery.
Hint
? You can use pipe() to pass updated messages from parent process to child process/ or from
child process to parent process.
? You can learn to use dtrace for profiling in this task.
Execution
main@ubuntu:~/Desktop$ g++ graph.cpp
main@ubuntu:~/Desktop$ ./a.out
Submission
? Due on: 23:59, 28 Feb 2023
? Plagiarism is strictly forbidden. Please note that TAs may ask you to explain the meaning of
your program to ensure that the codes are indeed written by yourself. Please also note that
we would check whether your program is too similar to your fellow students' code and
solutions available on the internet using plagiarism detectors.
? Late submission: A late submission within 15 minutes will not induce any penalty on your
grades. But 00:16 am-1:00 am: Reduced by 10%; 1:01 am-2:00 am: Reduced by 20%; 2:01
am-3:00 am: Reduced by 30% and so on. (e.g., Li Hua submitted a perfect attempt at 2:10
a.m. He will get (100+10 (bonus))*0.7=77p
? Any incorrect execution results by the following reasons are not arguable (Please double
check before submit)
? Compile error (Ensure no modification on header file)
? Incorrect file name/ path (Please strictly follow naming and path provided in instruction
and template)
? Successfully running program with incorrect result
Format guide
The project structure is illustrated below. You can also use ls command to check if your
structure is fine. Structure mismatch would cause grade deduction.
For this assignment, you don't need a specific folder for the bonus part. The source folder
should contain four files: graph.c, report.pdf . Please note that graph.h is not need to submit.
We limit your implementation within graph.c file, starting with "Task:" comments.
main@ubuntu:~/Desktop/Assignment_3_120010001$ ls
report.pdf graph.c
(One directory and one pdf.)
Please compress all files in the file structure root folder into a single zip file and name it using
your student ID as the code shown below and above, for example,
Assignment_3_120010001.zip. The report should be submitted in the format of pdf, together
with your source code. Format mismatch would cause grade deduction. Here is the sample step
for compressing your code.
main@ubuntu:~/Desktop$
zip -q -r Assignment_1_120010001.zip Assignment_3_120010001
main@ubuntu:~/Desktop$ ls
Assignment_1_120010001 Assignment_1_120010001.zip
Grading Rules
Program part 90'
You can test the correctness of your output using the following commands under directory.
? Execution with argument 0 will simply return grade
? Execution with argument 1 will return the test result of each case (1 for correct, 0 for wrong).
main@ubuntu:~/Desktop$ ./grader 1
Test result 1 0 0
main@ubuntu:~/Desktop$ ./grader 0
20
Task1_1 20p
Task1_2 30p
Task2 40p
Report 10p