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MrRissso [65]
2 years ago
13

What is the system that consists of nonproprietary hardware and software based on publicly known standards that allow third part

ies to create add-on products to plug into or interoperate with the system?
Computers and Technology
1 answer:
il63 [147K]2 years ago
5 0

Answer:

Open systems

Explanation:

Open systems are very different from Open Source applications or software, it should not be confused.

Open systems work with the blend of open software standards, portability, and interoperability. Computer systems that interoperate among multiple standards and vendors to ensure that computer resources (hardware and software) are not allotted to a particular vendor. Such computer systems are considered as open systems.

For instance, computer systems that run a Microsoft Windows OS can be considered as an Open system. This is because of their capability to run different versions of the Microsoft Windows OS on that particular computer system. More clearly, A computer with Windows 10 OS, can be used to install Windows 8 OS without any issue. That same computer system can run the Windows 7 OS. This makes the computer system and open system.

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7 0
2 years ago
. Write a code to define a linked list node using the above defined structure student (i.e. the data of the node is a student st
enyata [817]

Answer:

struct node{

   student data;

   node* next;

};

Explanation:

The above written is the segment of code is the structure of node of the linked list.The data of the node is type student.So the data of the node of the linked list will consist of the student details and the next is the pointer which holds the address of the next node.

8 0
3 years ago
Create a program which will input data into a pipe one character at a time. Count the number of characters as they are written i
LenKa [72]

Answer:

a)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

 

#define BUFSIZE 16

int main(){

   //pipe descriptors

   char msg[BUFSIZE];

   char buf[BUFSIZE];

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //pipe creation successfull

   //write four messages

   sprintf(msg , "apple");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "boy");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "cat");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "dog");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   //read

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   close(pipefd[0]);

   close(pipefd[1]);

   return 0;

}

(b)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<wait.h>

#define BUFSIZE 16

int main(){

   //pipe descriptors

   char msg[BUFSIZE];

   char buf[BUFSIZE];

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //pipe creation successfull

   //write four messages

   if(fork() == 0){

       //this is child

       //close unused write end

       close(pipefd[1]);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading first message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading second message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading third message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading fourth message. Content is %s\n", buf);

       close(pipefd[0]);

       //exit from child

       exit(EXIT_SUCCESS);

   }

   //this is parent process

   //close unused read end

   close(pipefd[0]);

   sprintf(msg , "apple");

   printf("This is parent process. Writing first message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "boy");

   printf("This is parent process. Writing second message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "cat");

   printf("This is parent process. Writing third message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "dog");

   printf("This is parent process. Writing fourth message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   close(pipefd[1]);

   //wait for child

   wait(NULL);

   return 0;

}

(c)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<sys/wait.h>

#include<sys/types.h>

#include<signal.h>

typedef struct sigaction Sigaction;

unsigned long long size = 0;

void alarmhandler(int sig){

   //alarm fired writing blocked

   printf("Write blocked after %llu characters\n", size);

   exit(EXIT_SUCCESS);

}

int main(){

   //pipe descriptors

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //install handler

   sigset_t mask , prev;

   sigemptyset(&mask);

   sigaddset(&mask , SIGALRM);

   sigprocmask(SIG_BLOCK , &mask , &prev);

   Sigaction new_action;

   sigemptyset(&new_action.sa_mask);

   new_action.sa_flags = SA_RESTART;

   new_action.sa_handler = alarmhandler;

   sigaction(SIGALRM , &new_action , NULL);

   sigprocmask(SIG_SETMASK , &prev, NULL);

   while(1){

     

       //print size on multiple of 4

       if(size != 0 && size % 1024 == 0){

           printf("%llu characters in pipe\n", size);

       }

       //reset previous alarm

       alarm(0);

       //set new alarm for 5 seconds

       alarm(5);

       //write to pipe one character

       write(pipefd[1], "A", sizeof(char));

       size++;

   }

   return 0;

}

Explanation:

Output for a, b, c are pasted accordingly

4 0
3 years ago
You just finished training a decision tree for spam classification, and it is gettingabnormally bad performance on both your tra
strojnjashka [21]

Answer:

Option B is the correct option.

Explanation:

The following answer is true because when the person completed our training of a decision tree and after the following presentation he getting not good working performance on both side i.e., test sets and during the training period. After the training there is no bug on the implementation of the presentation then, he has to increase the rate of the learning.

5 0
3 years ago
Which should you consider when selecting a highlighting color?
hjlf

Answer:contrast

Explanation:

To make sure it’s easier to read

5 0
3 years ago
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