What is happing to the parent process? - c++

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.

Answer 2

Your test for even numbers is wrong. You exclude bit 1, so any number greater than one will be selected. You effectively test for
if (buf & 0xfffffffe) ...
but, you should test instead for
if (!(buf & 1)) ...
or even
if ((buf % 2) == 0) ...

Answer 3

well, there are a couple of bugs in the code.
it looks like you want to achieve this: the child writes the random numbers and the parent waits for the child to terminate, reads the random numbers and only writes the even ones out to the second file.
however, you have used wait(), which is the wrong function to use. instead use waitpid() like this:
waitpid(pid, NULL, 0);
the other bug is in your code that writes only even numbers. I would write it like this:
if (!(buf & 1))
write(fd2, &buf1, sizeof(buf1));
that way, you dont need buf2 at all. in fact, in your current code, if the number is not even, it will still write a number to the file! It will just write the last even number that it read again.

Answer 4

Why are you using non-portable open/read/write when you could be using portable fopen/fread/fwrite? That makes no sense to me.
Let us suppose your sizeof int == 4. Suppose your first process writes 4 bytes, which your second process reads as two chunks of 2 bytes for whatever reason, thus read returns 2 here in your loop: while (read(fd, &buf, sizeof(buf)) == sizeof(buf)) Is 2 == sizeof (buf)? No. Perhaps your loop is terminating on the wrong condition...
fread solves this problem with two parameters: One for the size of the item and one for the number of items you want. for (size_t x = fread(&buf, sizeof buf, 1, file_ptr); x > 0; x = fread(&buf, sizeof buf, 1, file_ptr)) { /* ... */ }... "Upon successful completion, fread() shall return the number of elements successfully read which is less than nitems only if a read error or end-of-file is encountered." This guarantee doesn't exist for read. You might need to use x to determine how many items to write, since that tells you how many items were read.