initial commit

ex1/ex1.c

@@ -8,7 +8,19 @@
 
 int main(void)
 {
-    // Your code here
+
+    int x = 50;
+    int pid = fork();
+
+    if (pid == 0)
+    {
+        printf("Child value: %i\n", x);
+    }
+    else
+    {
+
+        printf("Parent value: %i\n", pid);
+    }
 
     return 0;
 }

ex2/ex2.c

@@ -1,5 +1,5 @@
-// Write a program that opens the text.txt file (with the `fopen()` library call) located in this directory 
-// and then calls `fork()` to create a new process. Can both the child and parent access the file descriptor 
+// Write a program that opens the text.txt file (with the `fopen()` library call) located in this directory
+// and then calls `fork()` to create a new process. Can both the child and parent access the file descriptor
 // returned by `fopen()`? What happens when they are written to the file concurrently?
 
 #include <stdio.h>
@@ -8,7 +8,21 @@
 
 int main(void)
 {
-    // Your code here 
-
+    FILE *fp;
+
+    fp = fopen("text.txt", "w+");
+
+    fprintf(fp, "%s %s", "Hello", "World");
+    fclose(fp);
+
+    int rc = fork();
+    if (rc == 0)
+    {
+        printf("I am the child, pid: %d\n", (int)getpid());
+    }
+    else
+    {
+        printf("I am the parent, pid: %d, pid of child: %d\n", (int)getpid(), rc);
+    }
     return 0;
 }

ex2/text.txt

@@ -0,0 +1 @@
+Hello World

ex3/ex3.c

@@ -9,7 +9,18 @@
 
 int main(void)
 {
-    // Your code here
+    // printf("goodbye (pid: %d)\n", (int)getpid());
+    int rc = fork();
 
+    // child process starts executing here
+    if (rc == 0)
+    {
+        printf("Child: hello\n");
+    }
+    else
+    {
+        int wc = waitpid(rc, NULL, 0);
+        printf("Parent: goodbye\n");
+    }
     return 0;
 }

ex4/ex4.c

@@ -1,6 +1,6 @@
 // Write a program that calls `fork()` and then calls some form of `exec()`
-// to run the program `/bin/ls`. Try a few variants of `exec()`, such as 
-// `execl()`, `execle()`, `execv()`, and others. Why do you think there 
+// to run the program `/bin/ls`. Try a few variants of `exec()`, such as
+// `execl()`, `execle()`, `execv()`, and others. Why do you think there
 // are so many variants of the same basic call?
 
 #include <stdio.h>
@@ -10,7 +10,11 @@
 
 int main(void)
 {
-    // Your code here    
+    printf("Hello World (pid: %d)\n", (int)getpid());
+
+    int rc = fork();
+
+    rc = execl("/bin/ls", "ls", NULL);
 
     return 0;
 }

ex5/ex5.c

@@ -1,7 +1,7 @@
 // Write a program that forks a child and creates a shared pipe
-// between the parent and child processes. Have the child write 
-// the three messages to the parent and have the parent print out 
-// the messages. 
+// between the parent and child processes. Have the child write
+// the three messages to the parent and have the parent print out
+// the messages.
 
 #include <stdio.h>
 #include <unistd.h>
@@ -10,13 +10,34 @@
 
 #define MSGSIZE 16
 
-char* msg1 = "hello world #1";
-char* msg2 = "hello world #2";
-char* msg3 = "hello world #3";
+char *msg1 = "hello world #1";
+char *msg2 = "hello world #2";
+char *msg3 = "hello world #3";
 
 int main(void)
 {
-    // Your code here
-
+    char inbuf[MSGSIZE];
+    int p[2], i;
+
+    if (pipe(p) < 0)
+        exit(1);
+
+    int rc = fork();
+    if (rc == 0)
+    {
+        printf("Child process\n");
+        write(p[1], msg1, MSGSIZE);
+        write(p[1], msg2, MSGSIZE);
+        write(p[1], msg3, MSGSIZE);
+    }
+    else
+    {
+        for (i = 0; i < 3; i++)
+        {
+            read(p[0], inbuf, MSGSIZE);
+            printf("% s\n", inbuf);
+        }
+    }
+
     return 0;
 }

ex6/ex6.c

@@ -20,7 +20,26 @@ and `clock_gettime()` should work just fine.
 
 int main()
 {
-    // Your code here
-
+    const char msg[] = "Hello World!";
+    int time;
+
+    for (int i = 0; i < number_iter; i++)
+    {
+        struct timespec start, end;
+
+        clock_gettime(CLOCK_MONOTONIC, &start); /* mark start time */
+
+        write(STDOUT_FILENO, msg, sizeof(msg) - 1);
+
+        clock_gettime(CLOCK_MONOTONIC, &end); /* mark the end time */
+
+        time += BILLION * (end.tv_sec - start.tv_sec) + end.tv_nsec - start.tv_nsec;
+        ;
+    }
+
+    time /= number_iter;
+
+    printf("average elapsed time = %d nanoseconds\n", time);
+
     return 0;
 }