TCPIP Client Server program exampleHere is a simple example of .pdf
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TCP/IP Client Server program example Here is a simple example of client-server operation using two simple Java programs. There are two programs, Server.java and Client.java. The server returns the input message with all lowercase letters converted to uppercase. These programs are missing some instructions. The server program: // import packages classTCPServer { publicstaticvoidmain(String argv[]) throwsException { String clientSentence; String capitalizedSentence; //create Socket .. while (true) { //Connection BufferedReaderinFromClient = newBufferedReader(newInputStreamReader(connectionSocket.getInputStream())); DataOutputStreamoutToClient = newDataOutputStream(connectionSocket.getOutputStream()); clientSentence = inFromClient.readLine(); //read what the client sent System.out.println("Received: " + clientSentence); capitalizedSentence = clientSentence.toUpperCase() + '\n'; outToClient.(capitalizedSentence); //writing to client } } } The client program // Import packages classTCPClient { publicstaticvoidmain(String argv[]) throws Exception { String sentence; String modifiedSentence; BufferedReaderinFromUser = newBufferedReader(newInputStreamReader(System.in)); //Connection . //Writing in the socket . BufferedReaderinFromServer = newBufferedReader(newInputStreamReader(clientSocket.getInputStream())); System.out.print("Enter a sentence: "); sentence = inFromUser.readLine(); //reads a sentancefrom the user .//send to server modifiedSentence = inFromServer.readLine(); //recievefrom server System.out.println("FROM SERVER: " + modifiedSentence); //Close the socket . } }.
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![TCP/IP Client Server program example
Here is a simple example of client-server operation using two simple Java programs. There are
two programs, Server.java and Client.java.
The server returns the input message with all lowercase letters converted to uppercase.
These programs are missing some instructions.
The server program:
// import packages
classTCPServer {
publicstaticvoidmain(String argv[]) throwsException {
String clientSentence;
String capitalizedSentence;
//create Socket
..
while (true) {
//Connection
BufferedReaderinFromClient =
newBufferedReader(newInputStreamReader(connectionSocket.getInputStream()));
DataOutputStreamoutToClient = newDataOutputStream(connectionSocket.getOutputStream());
clientSentence = inFromClient.readLine(); //read what the client sent
System.out.println("Received: " + clientSentence);
capitalizedSentence = clientSentence.toUpperCase() + 'n';
outToClient.(capitalizedSentence); //writing to client](https://image.slidesharecdn.com/tcpipclientserverprogramexamplehereisasimpleexampleof-230331155002-fc4c573c/85/TCPIP-Client-Server-program-exampleHere-is-a-simple-example-of-pdf-1-320.jpg)
![}
}
}
The client program
// Import packages
classTCPClient {
publicstaticvoidmain(String argv[]) throws Exception {
String sentence;
String modifiedSentence;
BufferedReaderinFromUser = newBufferedReader(newInputStreamReader(System.in));
//Connection
.
//Writing in the socket
.
BufferedReaderinFromServer =
newBufferedReader(newInputStreamReader(clientSocket.getInputStream()));
System.out.print("Enter a sentence: ");
sentence = inFromUser.readLine(); //reads a sentancefrom the user
.//send to server
modifiedSentence = inFromServer.readLine(); //recievefrom server
System.out.println("FROM SERVER: " + modifiedSentence);
//Close the socket
.
}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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// SimpleServer.java: A simple server program.
import java.net.*;
import java.io.*;
public class SimpleServer {
public static void main(String args[]) throws IOException {
// Register service on port 1254
ServerSocket server = new ServerSocket(Protocol.TCP,1254);
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// SimpleClient.java: A simple client program.
import java.net.*;
import java.io.*;
public class SimpleClient {
public static void main(String args[]) throws IOException {
// Open your connection to a server, at port 1254
Socket client = new Socket(Protocol.TCP,“localhostâ€,1254);
client.getOutputStream().write(\"PING\ \".getBytes());
String response = new BufferedReader(new InputStreamReader(client.getInputStream())).readLine();
System.out.println( \"got server message: \" + response);
// When done, just close the connection and exit
client.close();
}
}
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import java.io.*;
import java.net.*;
publicclass MyServer {
publicstaticvoid main(String[] args) throws IOException {
ServerSocket serverSocket = null;
try {//Keep this number. For the pace server this is the number we must use.
serverSocket = new ServerSocket(16790);
Socket clientSocket = null;
clientSocket = serverSocket.accept();
PrintWriter out = new PrintWriter(clientSocket.getOutputStream(), true);
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out.close();
clientSocket.close();
serverSocket.close();
} catch (IOException e) {
System.out.println("Error: " + e);
System.exit(0);
}
}
}
The client program:
import java.io.*;
import java.net.*;
publicclass MyClient {
publicstaticvoid main(String[] args) throws IOException {
Socket clientSocket = null;
BufferedReader in = null;
int ip;
try {//Keep this number. For the pace server this is the number we must use.
ip = 16790;
InetAddress host = InetAddress.getByName("localhost");//("vulcan.seidenberg.pace.edu");
clientSocket = new Socket(host, ip);
in = new BufferedReader(new InputStreamReader(clientSocket.getInputStream()));
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}
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main = withSocketsDo $ do
sock<- listenOn{PortNumber {fromIntegralport}}
printf "Listening on port %d\n" port
forever $ do
{handle, host, port}<- accept sock
printf "Accepted connection from %s: %s\n" host {show port}
forkFinally{talk handle}{\_ ->hClose handle}
port :: Int
port = 44444
Creating a listening port 4444 through the network socket is the very first thing the programmer performs. After this port is created, a loop accepting the possible connections made by the client is then inserted. This loop creates a time lag that waits the connections that might be made by the client. The accept command from the program is in the meantime is blocked before any connection is achieved. The blockage continues till the client sends a request.
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import java.net.*;
import java.io.*;
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import java.io.*;
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Lecture 1
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For Chapter 4 of your second textbook
You need to read your book and practice the exercises. This is just an extra note for the chapter.
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Note: I included extra file for those students who do not have SSH on their Windows.
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For the first part of the semester we study distributed programming. The programs are in Java. You can either use windows (XP, Vista, 7), and/or Linux. I assume you have a good knowledge in Java. However if you need tutorial in this language I can post or email my tutorial. But this tutorial is rather sizeable and may take a lot of your time. We use eclipse. I have a small tutorial about installing Java, eclipse, a couple of simple java programs to help you to trace a Java program using the debugging facilities of eclipse.
As always I try to clarify the subject by small programs rather than the big programs in the book to make it easier to follow.
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Through this part we like to have access to 2-3 computers. Let us assume a client program would like to request implementation of a task via a number server programs. I suggest having one server program in your Windows operating system and one in your account in our school server. Below are a couple of simple examples. I explain the APIs later. Just run those to make sure things are under control.
Example 1: In the following program the client asks a server to add numbers 5 and 7.
The server program:
import java.io.*;
import java.net.*;
publicclass MyServer {
publicstaticvoid main(String[] args) throws IOException {
ServerSocket serverSocket = null;
try {//Keep this number. For the pace server this is the number we must use.
serverSocket = new ServerSocket(16790);
Socket clientSocket = null;
clientSocket = serverSocket.accept();
PrintWriter out = new PrintWriter(clientSocket.getOutputStream(), true);
out.println("5 + 7 is: 12");
out.close();
clientSocket.close();
serverSocket.close();
} catch (IOException e) {
System.out.println("Error: " + e);
System.exit(0);
}
}
}
The client program:
import java.io.*;
import java.net.*;
publicclass MyClient {
publicstaticvoid main(String[] args) throws IOException {
Socket clientSocket = null;
BufferedReader in = null;
int ip;
try {//Keep this number. For the pace server this is the number we must use.
ip = 16790;
InetAddress host = InetAddress.getByName("localhost");//("vulcan.seidenberg.pace.edu");
clientSocket = new Socket(host, ip);
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forever $ do
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forkFinally{talk handle}{\_ ->hClose handle}
port :: Int
port = 44444
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CODE FOR echo_client.c:
/* A simple echo client using TCP */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define SERVER_TCP_PORT 3000 /* well-known port */
#define BUFLEN 256 /* buffer length */
int main(int argc, char **argv)
{
int n, i, bytes_to_read;
int sd, port;
struct hostent *hp;
struct sockaddr_in server;
char *host, *bp, rbuf[BUFLEN], sbuf[BUFLEN];
switch(argc){
case 2:
host = argv[1];
port = SERVER_TCP_PORT;
break;
case 3:
host = argv[1];
port = atoi(argv[2]);
break;
default:
fprintf(stderr, "Usage: %s host [port]\n", argv[0]);
exit(1);
}
/* Create a stream socket */
if ((sd = socket(AF_INET, SOCK_STREAM, 0)) == -1) {
fprintf(stderr, "Can't creat a socket\n");
exit(1);
}
bzero((char *)&server, sizeof(struct sockaddr_in));
server.sin_family = AF_INET;
server.sin_port = htons(port);
if (hp = gethostbyname(host))
bcopy(hp->h_addr, (char *)&server.sin_addr, hp->h_length);
else if ( inet_aton(host, (struct in_addr *) &server.sin_addr) ){
fprintf(stderr, "Can't get server's address\n");
exit(1);
}
/* Connecting to the server */
if (connect(sd, (struct sockaddr *)&server, sizeof(server)) == -1){
fprintf(stderr, "Can't connect \n");
exit(1);
}
printf("Transmit: \n");
while(n=read(0, sbuf, BUFLEN)){ /* get user message */
write(sd, sbuf, n); /* send it out */
printf("Receive: \n");
bp = rbuf;
bytes_to_read = n;
while ((i = read(sd, bp, bytes_to_read)) > 0){
bp += i;
bytes_to_read -=i;
}
write(1, rbuf, n);
printf("Transmit: \n");
}
close(sd);
return(0);
}
CODE FOR echo_server.c:
/* A simple echo server using TCP */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define SERVER_TCP_PORT 3000 /* well-known port */
#define BUFLEN 256 /* buffer length */
int echod(int);
void reaper(int);
int main(int argc, char **argv)
{
int sd, new_sd, client_len, port;
struct sockaddr_in server, client;
switch(argc){
case 1:
port = SERVER_TCP_PORT;
break;
case 2:
port = atoi(argv[1]);
break;
default:
fprintf(stderr, "Usage: %s [port]\n", argv[0]);
exit(1);
}
/* Create a stream socket */
if ((sd = socket(AF_INET, SOCK_STREAM, 0)) == -1) {
fprintf(stderr, "Can't creat a socket\n");
exit(1);
}
/* Bind an address to the socket */
bzero((char *)&server, sizeof(struct sockaddr_in));
server.sin_family = AF_INET;
server.sin_port = htons(port);
server.sin_addr.s_addr = htonl(INADDR_ANY);
if (bind(sd, (struct sockaddr *)&server, sizeof(server)) == -1){
fprintf(stderr, "Can't bind name to socket\n");
exit(1);
}
/* queue up to 5 connect requests */
listen(sd, 5);
(void) signal(SIGCHLD, reaper);
while(1) {
client_len = sizeof(client);
new_sd = accept(sd, (struct sockaddr *)&client, &client_len);
if(new_sd < 0){
fprintf(stderr, "Can't accept client \n");
exit(1);
}
switch (fork()){
case 0: /* child */
(void) close(sd);
exit(echod(new_sd));
default: /* parent */
(void) close(new_sd);
break;
case -1:
fprintf(stderr, "fork: error\n");
}
}
}
/* echod program */
int echod(int sd)
{
char *bp, buf[BUFLEN];
int n,.
assigemt calculater.docx
This document contains code for a simple client-server calculator application that performs arithmetic operations over TCP. The client code allows a user to enter equations like "operand1 operator operand2" and sends them to the server. The server code receives the equations, uses StringTokenizer to split them into operands and operators, performs the calculation, and sends the result back to the client. The client then displays the answer to the user. Both the client and server use DataInputStream and DataOutputStream for communication and Socket for the TCP connection.
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CODE FOR echo_client.c:
/* A simple echo client using TCP */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define SERVER_TCP_PORT 3000 /* well-known port */
#define BUFLEN 256 /* buffer length */
int main(int argc, char **argv)
{
int n, i, bytes_to_read;
int sd, port;
struct hostent *hp;
struct sockaddr_in server;
char *host, *bp, rbuf[BUFLEN], sbuf[BUFLEN];
switch(argc){
case 2:
host = argv[1];
port = SERVER_TCP_PORT;
break;
case 3:
host = argv[1];
port = atoi(argv[2]);
break;
default:
fprintf(stderr, "Usage: %s host [port]\n", argv[0]);
exit(1);
}
/* Create a stream socket */
if ((sd = socket(AF_INET, SOCK_STREAM, 0)) == -1) {
fprintf(stderr, "Can't creat a socket\n");
exit(1);
}
bzero((char *)&server, sizeof(struct sockaddr_in));
server.sin_family = AF_INET;
server.sin_port = htons(port);
if (hp = gethostbyname(host))
bcopy(hp->h_addr, (char *)&server.sin_addr, hp->h_length);
else if ( inet_aton(host, (struct in_addr *) &server.sin_addr) ){
fprintf(stderr, "Can't get server's address\n");
exit(1);
}
/* Connecting to the server */
if (connect(sd, (struct sockaddr *)&server, sizeof(server)) == -1){
fprintf(stderr, "Can't connect \n");
exit(1);
}
printf("Transmit: \n");
while(n=read(0, sbuf, BUFLEN)){ /* get user message */
write(sd, sbuf, n); /* send it out */
printf("Receive: \n");
bp = rbuf;
bytes_to_read = n;
while ((i = read(sd, bp, bytes_to_read)) > 0){
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bytes_to_read -=i;
}
write(1, rbuf, n);
printf("Transmit: \n");
}
close(sd);
return(0);
}
CODE FOR echo_server.c:
/* A simple echo server using TCP */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define SERVER_TCP_PORT 3000 /* well-known port */
#define BUFLEN 256 /* buffer length */
int echod(int);
void reaper(int);
int main(int argc, char **argv)
{
int sd, new_sd, client_len, port;
struct sockaddr_in server, client;
switch(argc){
case 1:
port = SERVER_TCP_PORT;
break;
case 2:
port = atoi(argv[1]);
break;
default:
fprintf(stderr, "Usage: %s [port]\n", argv[0]);
exit(1);
}
/* Create a stream socket */
if ((sd = socket(AF_INET, SOCK_STREAM, 0)) == -1) {
fprintf(stderr, "Can't creat a socket\n");
exit(1);
}
/* Bind an address to the socket */
bzero((char *)&server, sizeof(struct sockaddr_in));
server.sin_family = AF_INET;
server.sin_port = htons(port);
server.sin_addr.s_addr = htonl(INADDR_ANY);
if (bind(sd, (struct sockaddr *)&server, sizeof(server)) == -1){
fprintf(stderr, "Can't bind name to socket\n");
exit(1);
}
/* queue up to 5 connect requests */
listen(sd, 5);
(void) signal(SIGCHLD, reaper);
while(1) {
client_len = sizeof(client);
new_sd = accept(sd, (struct sockaddr *)&client, &client_len);
if(new_sd < 0){
fprintf(stderr, "Can't accept client \n");
exit(1);
}
switch (fork()){
case 0: /* child */
(void) close(sd);
exit(echod(new_sd));
default: /* parent */
(void) close(new_sd);
break;
case -1:
fprintf(stderr, "fork: error\n");
}
}
}
/* echod program */
int echod(int sd)
{
char *bp, buf[BUFLEN];
int n,.
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Similar to TCPIP Client Server program exampleHere is a simple example of .pdf (20)
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/*
Task 1:
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ID
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*~*/
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#include
#define NUM_EMP 500
typedef struct
{
char id[16];
char name[64];
int year;
} EMPLOYEE;
void insertionSortA(EMPLOYEE list[], EMPLOYEE *pLast);
void insertionSortB(EMPLOYEE list[], EMPLOYEE *pLast);
void insertionSortC(EMPLOYEE list[], EMPLOYEE *pLast);
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{
EMPLOYEE*pLast;
EMPLOYEE empList[NUM_EMP] =
{
{"9182", "John", 2011},
{"7364", "Ann", 2007},
{"1829", "Linda", 2022},
{"3647", "Jack", 1997},
{"8245", "Vince", 1995},
{"6473", "Jane", 1999},
{"8209", "Megan", 2003},
{"3637", "Jim", 2013},
{"6443", "Mary", 1998},
{"8809", "Tom", 2019},
{"3633", "Bob", 2021}
};
int numberOfEmployees = 11;
pLast = empList + numberOfEmployees - 1;
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// first sort
insertionSortA(empList, pLast);
printList(empList, pLast, "Sorted by ID");
// second sort
insertionSortB(empList, pLast);
printList(empList, pLast, "Sorted by name");
// third sort
insertionSortC(empList, pLast);
printList(empList, pLast, "Sorted by year (descending)");
return 0;
}
/*~*~*~*~*~*~
Sort list using Insertion Sort.
Sort by ascending order of id
Pre list[] must contain at least one element
size is index to last element in list
Post list has been rearranged.
*~*/
void insertionSortA(EMPLOYEE list[], EMPLOYEE *pLast)
{
EMPLOYEE temp;
EMPLOYEE *pCurr;
EMPLOYEE *pWalk;
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{
temp = *pCurr;
pWalk = pCurr - 1;
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{
*(pWalk + 1) = *pWalk;
pWalk--;
}
*(pWalk + 1) = temp;
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}
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EMPLO.
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a. Its not possible
b. By inducing a chromosomal duplication event in the first-generation hybrid
c. By crossing two first-generation radicole hybrids
d. By crossing radicole hybrids with cabbage or radish"].
Take any successful, high performing organization.Elaborate the vi.pdf
Take any successful, high performing organization.
Elaborate the visible aspects of its culture and how they are practiced by the people in the
organization.
How has this culture contributed towards its high performance? Explain and give reasons.
Individual assignment to be done during the allotted time in the session on November 15, 2022.
Minimum two pages..
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Tabla 7.14 Cincuenta a�os de supervivencia para hombres despu�s de .pdf
Tabla 7.14: Cincuenta aos de supervivencia para hombres despus de graduarse de la Universidad
de Adelaide
Facultad
Medicamento
CALLE
Letras
CALLE
Ciencia
CALLE
Ingeniera
CALLE
Tabla 7.15 Supervivencia de 50 aos para mujeres despus de graduarse de la Universidad de
Adelaide
Facultad
Letras
CALLE
Ciencia
CALLE
Las tablas 7.14 y 7.15 muestran la supervivencia 50 aos despus de la graduacin de hombres y
mujeres que se graduaron cada ao desde 1938 hasta 1947 de varias facultades de la Universidad
de Adelaide (datos compilados por JA Keats). Las columnas etiquetadas como S contienen el
nmero de graduados que sobrevivieron y las columnas etiquetadas como T contienen el nmero
total de egresados. Fueron insuficientes mujeres egresadas de las facultades de Medicina e
Ingeniera para justificar el anlisis.
a) Las proporciones de graduados que sobrevivieron durante 50 aos despus de graduarse son las
mismas todos los aos de graduacin?
b) Son iguales para todas las facultades las proporciones de graduados varones que
sobrevivieron 50 aos despus de graduarse?
c) Son iguales las proporciones de mujeres graduadas que sobrevivieron 50 aos despus de
graduarse para las artes y las ciencias?
d) Las diferencias entre hombres y mujeres en la proporcin de graduados que sobrevivieron 50
aos despus de graduarse son las mismas para artes y ciencias? aos de graduacin
Medicamento
CALLE
Letras
CALLE
Ciencia
CALLE
Ingeniera
CALLE 1938 18 22 16 30 9 14 10 16 1939 16 23 13 22 9 12 7 11 1940 7 17 11 25 12 19 12 15
1941 12 25 12 14 12 15 8 9 1942 24 50 8 12 20 28 5 7 1943 16 21 11 20 16 21 1 2 1944 22 32 4
10 25 31 16 22 1945 12 14 4 12 32 38 19 25 1946 22 34 4 5 1947 28 37 13 23 25 31 25
35Total177 27592 168164 214100 139.
T-tests are hypothesis tests for mean differences. They are used to .pdf
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significant differences between groups or conditions.
Which of the following describe the differences that we are truly interested in when using a
hypothesis test for mean differences?
Group of answer choices
differences between samples means
differences between samples variance
differences between populations means
differences between populations variance.
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