Question

The task here is :This is basic C program which must contains two processes.First one writes in a file 60 random numbers.The seconds that needs to be written has to read the file and write only the even number into another new file.So I have this code and the parent process that needs to write the even numbers to a file its not working properly.Also my question is about the structure of the code.Is this the best way to describe the processes.

#include <stdio.h>
#include <fcntl.h> 

int main() {
    int pid, fd, buf, i, fd2, buf2;
    pid = fork();

    switch (pid) {
    case 0: {
        printf("\n Child process is writing numbers. \n");
        fd = open("file.dat", O_RDWR | O_CREAT, 0644);

        for (i = 0; i < 60; i++) {
            buf = rand() % 20 - 10;
            printf("\n Writing number %d %d ", i + 1, buf);
            write(fd, &buf, sizeof(buf));
        }

        close(fd);
        break;
    }

    case -1: {
        printf("\n Fork error! \n");
    }

    default: {
        wait(pid);
        printf("\n Parent process is Copying numbers. \n");
        fd = open("file.dat", O_RDONLY);
        fd2 = open("file_output.dat", O_RDWR | O_CREAT, 0644);

        while (read(fd, &buf, sizeof(buf)) == sizeof(buf)) {
            if (buf & ~1)
                (buf2 = buf);
            write(fd2, &buf2, sizeof(buf2));
            printf("Writing number in file_output.dat %d \n", buf2);
        }

        close(fd2);
        close(fd);
    }
    }
    return 0;
}

Answer 1

Your condition is wrong:

    if (buf &~ 1) (buf2=buf);
    write(fd2, &buf2, sizeof(buf2));

You write the number out regardless, but if it’s odd (excepting that you need to fix the even/odd test) then you don’t update buf2. If the first number is odd, buf2 is uninitialised.

You need this:

    if (buf % 2 != 0) continue;
    write(fd2, &buf, sizeof(buf));
    printf("Writing number in file_output.dat %d \n", buf);

Note that this might not correctly test negative even numbers. You could test abs(buf) % 2 != 0. The only number that might not be handled would be INT_MIN.