Merge pull request #7 from linomp/dev

Dev

.vscode/settings.json

@@ -0,0 +1,6 @@
+{
+    "files.associations": {
+        "*html": "html",
+        "unistd.h": "c"
+    }
+}

ostep/Makefile

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+CC=gcc
+CCOPTS=#-Wall -Wextra
+LIBS=
+
+SRCS=$(wildcard *.c)
+TARGETS=$(SRCS:.c=)
+
+.PHONY: all clean pristine
+
+all: $(TARGETS)
+
+clean:
+	rm -f $(TARGETS) & rm -f *.output
+
+pristine: clean all
+
+%: %.c
+	$(CC) $(CCOPTS) -o $@ $< $(LIBS)

ostep/ex5_1.c

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+#include <stdio.h>
+#include <stdlib.h>
+#include <sys/wait.h>
+#include <unistd.h>
+
+int main(int argc, char* argv[]) {
+    int* x = (int*)malloc(sizeof(int));
+
+    int rc = fork();
+
+    if (rc == 0) {
+        for (int i = 0; i <= 10; i++) {
+            printf("child | x=%d\n", (*x)++);
+            usleep(1);
+        }
+
+    } else {
+        for (int i = 0; i <= 10; i++) {
+            printf("parent| x=%d\n", (*x)++);
+            usleep(1);
+        }
+
+        wait(NULL);
+    }
+
+    return 0;
+}
+
+/*
+Q: Write a program that calls fork(). Before calling fork(), have the
+main process access a variable (e.g., x) and set its value to something (e.g.,
+100). What value is the variable in the child process? What happens to the
+variable when both the child and parent change the value of x?
+
+A:
+No race condition happens because they don't share memory actually.
+Child process gets its own "private memory" (copy of parent's address space)
+Therefore parent & child modify their own copy of x, indepedently.
+
+Things would break if they were using shared memory.
+
+note: Even if x is a pointer, nothing changes!
+*/

ostep/ex5_2.c

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+#include <assert.h>
+#include <fcntl.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <sys/wait.h>
+#include <unistd.h>
+
+int main(int argc, char* argv[]) {
+    int fd = open("./ex5_2.output", O_CREAT | O_WRONLY | O_TRUNC, S_IRWXU);
+    int rc = fork();
+
+    if (rc == 0) {
+        char* msg = "msg from child\n";
+
+        for (int i = 0; i < 5; i++) {
+            write(fd, msg, strlen(msg));
+            usleep(100);
+        }
+
+    } else {
+        char* msg = "msg from parent\n";
+        write(fd, msg, strlen(msg));
+
+        for (int i = 0; i < 5; i++) {
+            write(fd, msg, strlen(msg));
+            usleep(200);
+        }
+
+        wait(NULL);
+        close(fd);
+    }
+
+    return 0;
+}
+
+/*
+Q: Write a program that opens a file (with the open() system call)
+and then calls fork() to create a new process. Can both the child
+and parent access the file descriptor returned by open()? What
+happens when they are writing to the file concurrently, i.e., at the
+same time?
+
+A: the order in which the content will be written to the file is
+undeterministic. I needed to introduce loops & delays to force the race
+condition to happen, but it was clear
+*/

ostep/ex5_3.c

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+#include <assert.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <sys/wait.h>
+#include <unistd.h>
+
+int main() {
+    int pipe_fds[2];
+    assert(pipe(pipe_fds) >= 0);
+
+    int pipe_read_fd = pipe_fds[0];
+    int pipe_write_fd = pipe_fds[1];
+
+    int rc = fork();
+    assert(rc >= 0);
+
+    if (rc == 0) {
+        sleep(2);  // sleep to prove that nothing happens in parent until child
+                   // writes to pipe
+        char* msg = "msg from child!";
+        write(pipe_write_fd, msg, strlen(msg));
+    } else {
+        char buffer[200];
+        printf("parent blocked on read() ...\n");
+        read(pipe_read_fd, buffer, 200);
+        printf("parent received: “%s”\n", buffer);
+        printf("parent terminating\n");
+    }
+
+    return 0;
+}
+
+/*
+Q: Write another program using fork(). The child process should
+print “hello”; the parent process should print “goodbye”. You should
+try to ensure that the child process always prints first; can you do
+this without calling wait() in the parent?
+
+A: can be achieved using a pipe. parent blocks on read() until child writes to
+it.
+*/

ostep/ex5_8.c

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+#include <assert.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <sys/wait.h>
+#include <unistd.h>
+
+int main() {
+    int status;
+
+    int pipe_fds[2];
+    assert(pipe(pipe_fds) >= 0);
+
+    int pipe_read_fd = pipe_fds[0];
+    int pipe_write_fd = pipe_fds[1];
+
+    pid_t rc1 = fork();
+    assert(rc1 >= 0);
+
+    if (rc1 == 0) {
+        char* out_buffer = "msg from child 1!";
+
+        printf("child 1 is writing...\n");
+        write(pipe_write_fd, out_buffer, strlen(out_buffer));
+
+        close(pipe_write_fd);
+        printf("child 1 terminating...\n");
+
+        exit(0);
+    }
+
+    pid_t rc2 = fork();
+    assert(rc2 >= 0);
+
+    if (rc2 == 0) {
+        printf("child 2 blocked on read()...\n");
+
+        char in_buffer[200];
+        read(pipe_read_fd, in_buffer, 200);
+        printf("child 2 finished reading...\n");
+
+        char out_buffer[400];
+        sprintf(out_buffer, "%s || msg from child 2!", in_buffer);
+
+        printf("child 2 is writing...\n");
+        write(pipe_write_fd, out_buffer, strlen(out_buffer));
+
+        close(pipe_write_fd);
+        printf("child 2 terminating...\n");
+
+        exit(0);
+    }
+
+    // Parent process
+
+    printf("parent waiting for child 1 (pid: %d) ...\n", rc1);
+    waitpid(rc1, &status, 0);
+    printf("parent waiting for child 2 (pid: %d) ...\n", rc1);
+    waitpid(rc2, &status, 0);
+
+    char buffer[200];
+    read(pipe_read_fd, buffer, 200);
+    printf("parent received: “%s”\n", buffer);
+
+    close(pipe_write_fd);
+    printf("parent terminating...\n");
+
+    return 0;
+}
+
+/*
+Q: Write a program that creates two children, and connects the standard output
+of one to the standard input of the other, using the pipe() system call.
+*/