hi there i have a work about a programme which is like one child should print a number into a text file and the second child should take that number to print it onto screen simultaneously. but my code is work like first child finishes to printing the numbers 0 through 9 and then second child starts to read them onto screen. so i guess its a synchronization issue. here is my simple code ;
#include <stdio.h> /* basic I/O routines. */
#include <unistd.h> /* define fork(), etc. */
#include <sys/types.h> /* define pid_t, etc. */
#include <sys/wait.h> /* define wait(), etc. */
#include <signal.h> /* define signal(), etc. */
#include <pthread.h>
void write_num(void);
void print_screen(void);
//void catch_child(int);
int main(int argc, char* argv[]) {
int i, result, pid;
pid = fork(); /* creates a new process, you only want the parent to spawn children? */
switch (pid) {
case -1:
/* fork error */
printf("Error occured with fork()\n");
exit(1);
case 0:
/* child process */
write_num();
exit(0);
default:
/* parent process*/
{
//wait(&pid);
pid = fork(); /* fork new children here */
switch(pid) {
case -1:
printf("Error occured with fork()\n");
exit(1);
case 0:
print_screen();
exit(0);
default:
break;
}
}
}
wait(&pid);
execl("/usr/bin/killall","killall","tail",(char *) 0);
return 0;
}
void write_num(void){
FILE* fptr;
int i;
fptr=fopen("textfile.txt","w");
for(i=0; i<10; i++){
fprintf(fptr,"%d\n",i);
fflush(stdout);
sleep(1);
}
}
void print_screen(void){
execl("/usr/bin/tail","tail","-f","./textfile.txt",(char *) 0);
sleep(1);
}
/* first, here is the code for the signal handler
void catch_child(int sig_num)
{
when we get here, we know there's a zombie child waiting
int child_status;
wait(&child_status);
printf("child exited.\n");
}*/
by the way, in Ubuntu i used to compile with gcc -o process process.c -lpthread.
I will be appreciated if you can help.
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