簡易的C/S系統(實現兩個數的和)

標籤：

//Client:
#include <string.h>#include <sys/socket.h>#include <stdio.h>#include <netinet/in.h>int main(int argc, char **argv){        int sockfd;        struct sockaddr_in address;        int result;        int num1 = 25, num2 = 30, num3 = 0;        printf("請輸入兩個整數");        scanf("%d%d", &num1, &num2);        sockfd = socket(AF_INET,SOCK_STREAM, 0);        address.sin_family = AF_INET;        address.sin_addr.s_addr = inet_addr("127.0.0.1");        address.sin_port = 6666;        result = connect(sockfd, (struct sockaddr *) &address, sizeof(address));        if(result == -1){                perror("串連失敗了");                return 1;        }        write(sockfd, &num1, sizeof(num1));        write(sockfd, &num2, sizeof(num2));        read(sockfd, &num3, sizeof(num3));        printf("兩個數的和是%d\n", num3);        close(sockfd);        return 0;}    

//Server
include<stdio.h>#include<string.h>#include<sys/socket.h>#include<netinet/in.h>int main(int argc, int **argv){        int server_sockfd, client_sockfd;        struct sockaddr_in server_address;        struct sockaddr_in client_address;        //struct sockaddr_in *client_address2;        server_sockfd = socket(AF_INET, SOCK_STREAM, 0);        server_address.sin_family = AF_INET;        server_address.sin_addr.s_addr = inet_addr("127.0.0.1");        server_address.sin_port = 6666;        bind(server_sockfd, (struct sockaddr *) &server_address, sizeof(server_address));        listen(server_sockfd, 10);        int client_len;        //char client_ip[100], client_port[100];        while(1){                int num1, num2, num3;                printf("伺服器等待訊息\n");                client_len = sizeof(client_address);                client_sockfd = accept(server_sockfd, (struct sockaddr *) &client_address, (socklen_t *__restrict) &client_len);                //sprintf(client_ip, inet_ntoa(client_address.sin_addr));                //client_address2 = (struct sockaddr_in *)&client_address;                //printf("你所串連的主機的ip地址為: %s, 連接埠號碼是：\n", inet_ntoa(client_address2->sin_addr));                read(client_sockfd, &num1, sizeof(num1));                printf("第一個數是: %d\n", num1);                read(client_sockfd, &num2, sizeof(num2));                printf("第二個數是: %d\n", num2);                num3 = num1 + num2;                write(client_sockfd, &num3, sizeof(num3));                close(client_sockfd);        }        return 0;}

 

簡易的C/S系統(實現兩個數的和)