Finish Lab 3
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17
Lab3/examples/1-no_join-fixed.c
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Lab3/examples/1-no_join-fixed.c
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/**To compile, don't forget to add -lpthread. Might not work without that */
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#include<pthread.h>
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#include<stdio.h>
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void* run (void* arg){
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printf("Hello from run\n");
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return NULL;
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}
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int main(){
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pthread_t thread; // variable to store the reference to the thread
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pthread_create(&thread, NULL, &run, NULL);
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printf("In main"); // This section will be executed in parallel
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pthread_join(thread,NULL); // necessary for waiting for the thread to finish
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printf("In main 2");
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return 0;
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}
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Lab3/examples/2-sleep no_join-fixed.c
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Lab3/examples/2-sleep no_join-fixed.c
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#include<pthread.h>
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#include<stdio.h>
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#include<stdlib.h>
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#include<unistd.h>
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#include<time.h>
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#include<sys/time.h>
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void* run (void* arg){
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printf("Hello from run\n");
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return NULL;
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}
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int main(){
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pthread_t thread;
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pthread_create(&thread, NULL, &run, NULL);
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// While the sleep() function will appear to provide synchronization,
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// it is incorrect, and if used for this purposes instead of join
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// or synchronization mechanism - the grade for the assignment will be
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// reduced
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sleep(2);
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printf("Back in main");
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return 0;
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}
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24
Lab3/examples/3-losing_track_of_threads-fixed.c
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Lab3/examples/3-losing_track_of_threads-fixed.c
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#include<pthread.h>
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#include<stdio.h>
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#include<stdlib.h>
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#include<unistd.h>
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#include<time.h>
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#include<sys/time.h>
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void* run (void* arg){
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sleep(10-(int)arg);
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printf("Hello from run\n");
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}
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int main(){
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pthread_t thread[10];
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int i = 0;
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for (i = 0 ; i < 10; ++i)
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pthread_create(&thread[i], NULL, &run,(void*)i);
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// If you plan to use the results of all the threads, consider
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// using join for all of the threads
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for (i = 0 ; i < 10; ++i)
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pthread_join(thread[i],NULL);
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printf("In main");
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}
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Lab3/examples/4-passing_data_wrongly.c
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Lab3/examples/4-passing_data_wrongly.c
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#include<pthread.h>
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#include<stdio.h>
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void* run (void* arg){
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int i = (int) arg;
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printf("Hello from run, arg is %i\n",i);
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}
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int main(){
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pthread_t thread[10];
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int i = 0;
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for (i = 0 ; i < 10; ++i)
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pthread_create(&thread, NULL, &run,(void*)i);
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// Notice how warnings are generated. This can be resolved by properly
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// allocating space for the thread parameters.
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// Note, that if you are using stack variable, the values might be corrupted
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for (i = 0 ; i < 10; i++){
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pthread_join(thread[i],NULL);
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}
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printf("In main");
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}
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Lab3/examples/4-passing_data_wrongly_fixed.c
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Lab3/examples/4-passing_data_wrongly_fixed.c
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#include<pthread.h>
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#include<stdio.h>
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#include <stdlib.h>
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typedef struct thread_args{
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char letter;
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int id;
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}Args;
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void* run (void* arg){
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Args* argg = (Args*) arg;
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int i = argg->id;
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printf("Hello from run %d\n", i);
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return NULL;
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}
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int main(int argc, char** argv){
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pthread_t thread[10];
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int i = 0;
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for (i = 0 ; i < 10; i++){
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Args* argg = (Args*) (malloc(sizeof(Args)));
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argg->letter = 'q';
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argg->id = i;
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pthread_create(&(thread[i]), NULL, &run,(void*)argg);
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}
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for (i = 0 ; i < 10; i++){
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pthread_join(thread[i],NULL);
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}
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// Free the resources somewhere, for long running program. Or let the OS handle that
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printf("In main");
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return 0;
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}
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Lab3/examples/5-pthread_detach.c
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Lab3/examples/5-pthread_detach.c
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#include<pthread.h>
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#include<stdio.h>
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void* run (void* arg){
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printf("Hello from run\n");
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return NULL;
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}
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int main(){
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pthread_t thread;
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pthread_create(&thread, NULL, &run, NULL);
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pthread_detach(thread);
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// This command will make thread detached. This means the resources will be released
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// upon the thread's completion - calling return
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// However, detached threads cannot be joined, which means if you care about the result
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// of the thread execution - you should not be using pthread_detach
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printf("In main");
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return 0;
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}
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Lab3/main.c
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Lab3/main.c
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#include<stdio.h>
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#include<sys/time.h>
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#include<stdlib.h>
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#include<unistd.h>
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#include<string.h>
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/**
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* Question 1: Run the program with the problem size of 1000 and 10 threads, what is the approximate speedup you are achieving?
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*
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* Answer 1: On my home machine single threading took 0.640239 seconds and multithreading took 0.194644 seconds.
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* On the lab machine single threading took 0.697472 seconds and multithreading took 0.114415 seconds.
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* This means that the my home machine is 3.29 times faster and the lab machine is 6.10 times faster.
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*
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* Question 2: Is there a problem size / number of threads combination that slows down the computation process? Why do you think it is happening?
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*
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* Answer 2: There is a problem with having too many threads for the number of hardware threads available. Increasing past this only
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* increases the overhead of creating and managing the threads. This is because the threads are not running in parallel and are instead
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* being switched between by the OS. At lower matrix sizes the cost of creating threads and managing them is greater than the cost of
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* just doing the computation in a single thread, so any combination where threads > the number of hardware threads will be slower, and
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* as the matrix size approaches 1, than the greater effect thread creation and management will have on the speed of the program.
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*
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* Question 3: What is the minimum size of the problem that benefits from creating an extra thread?
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*
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* Answer 3: The lowest size on my home machine that consistently benefited from an extra thread was 150, but this can change depending
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* on the specifications of the machine.
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*
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* Question 4: Does using the threads always improve execution duration?
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*
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* Answer 4: No, as the number of threads increases past the number of hardware threads available the execution duration increases
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* due to managing the threads, as well as low size matrices where the cost of creating and managing threads is greater than the
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* cost of just doing the computation in a single thread.
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*
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* Question 5: Guesstimate and comment on the nature of growth of the speedup with the number of threads – is it linear, exponential, are there any limits?
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*
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* Answer 5: The speedup is linear up to the number of hardware threads available, given a large enough matrix size.
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* Given a matrix of size x, the speedup of using y threads is approximately x/y, up to the number of hardware threads available.
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* After this the speedup will decrease as the number of threads increases, due to the overhead of creating and managing threads.
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**/
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#include <stdio.h>
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#include <sys/time.h>
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#include <time.h>
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#include <stdlib.h>
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#include <string.h>
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#include <pthread.h>
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#define MAXN 5
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@ -82,6 +116,23 @@ void *multiply_matrices_threaded(void *threadParams) {
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* threading capacity and parallelize the computation in such a
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* way that a thread computes result per one or more rows
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*/
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ThreadParams *t = (ThreadParams *) threadParams;
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int N = t->size;
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int row = t->row_index;
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int column = 0;
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int temp_result = 0;
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while (row < N) {
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column = 0;
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while (column < N) {
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temp_result = 0;
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for (int i = 0; i < t->size; i++) {
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temp_result = temp_result + t->first_array[row][i] * t->second_array[i][column];
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}
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t->result[row][column] = temp_result;
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column = column + 1;
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}
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row += t->max_threads;
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}
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}
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int main(int argc, char **argv) {
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* Write your code to create and use max_threads here, such that the threaded_result
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* is populated with the result of the computation.
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*/
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thr->result = threaded_result;
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pthread_t threads[max_threads];
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for (int i = 0; i < max_threads; i++) {
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ThreadParams *params = (ThreadParams *) malloc(sizeof(ThreadParams));
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params->first_array = array1;
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params->second_array = array2;
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params->result = threaded_result;
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params->row_index = i;
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params->size = size;
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params->max_threads = max_threads;
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pthread_create(&threads[i], NULL, &multiply_matrices_threaded, (void *) params);
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}
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for (int i = 0; i < max_threads; i++) {
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pthread_join(threads[i], NULL);
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}
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gettimeofday(&end, NULL);
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//The next line is inspired by https://linuxhint.com/gettimeofday_c_language/
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microseconds = (end.tv_sec * 1000000 + end.tv_usec) - (begin.tv_sec * 1000000 + begin.tv_usec);
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@ -129,7 +196,7 @@ int main(int argc, char **argv) {
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if (check_if_matrices_differ(result, threaded_result, size) != 0) {
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printf("Threaded result differ from single core computation, error\n");
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exit(0);
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exit(1);
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}
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return 0;
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}
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9
Lab3/try-for-error.sh
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Lab3/try-for-error.sh
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#!/bin/bash
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cond=0
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while [ $cond -eq 0 ]
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do
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./build/Lab3 10 10 &> output.txt
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cond=$?
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echo $cond
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done
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