aaa.c:
#include<stdio.h>
#include<unistd.h>
int main(void)
{
int i=0;
for(i=0;i<100;i++)
{
printf("%c", 'A');
usleep(200); //sleep in microsecs
}
return 0;
}
bbb.c
#include<stdio.h>
#include<unistd.h>
int main(void)
{
int i=0;
for(i=0;i<100;i++)
{
printf("%c", 'B');
usleep(200);
}
return 0;
}
ccc.c
#include<stdio.h>
int main(void)
{
char arr[2001];
scanf("%s",arr);
int i=0;
int size=sizeof(arr)/sizeof(int);
char c=arr[0];
int count=0;
for(i=0;i<size;i++)
{
if(c==arr[i])
{
count++;
}
else
{
printf("%c %d,",c,count);
c=arr[i];
count=1;
}
}
printf("\n");
return 0;
}
xyz.c
#include<fcntl.h>
#include<stdio.h>
#include<unistd.h>
#include<sys/stat.h>
#include<sys/wait.h>
int main(int argc, char *argv[])
{
int mypipe[2];
int child1=1,child2=1,child3=1;
pipe(mypipe);
child1=fork();
child2=fork();
child3=fork();
if(child1==0)
{
close(mypipe[0]);
dup2(mypipe[1],1);
close(mypipe[1]);
execl("aaa","aaa",NULL);
perror("cannot start aaa");
return 1;
}
if(child2==0)
{
close(mypipe[0]);
dup2(mypipe[1],1);
close(mypipe[1]);
execl("bbb","bbb",NULL);
perror("cannot start bbb");
return 1;
}
if(child3==0)
{
close(mypipe[1]);
dup2(mypipe[0],0);
close(mypipe[0]);
execl("ccc","ccc",NULL);
perror("cannot start ccc");
return 1;
}
return 0;
}
output:
A 200,B 200,
Note: The order of the output may differ based on the architecture of the CPUs.
Suppose we have three programs aaa, bbb, ccc which are already developed, compiled and ready to...
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#include <unistd.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <string.h>
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