C on LINUX - Multiple child proces exec unusual number of times - c

Question

I'm trying to create n = 10 child processes and make its execute a peace of code ..
However it creates 14 child processes indifferent of n.
Why is that?
This is the sample code :
#include <stdlib.h>
#include <stdio.h>
int main()
{
printf("It worked! ");
return 0;
}
And this is the main program:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <errno.h>
#include <unistd.h>
#include <sys/wait.h>
int main (int argc, char *argv[])
{
int n = 10;
pid_t pid;
int status = 0;
int fd2[2];
int i = 0;
while (i < n)
{
/*create the pipe */
if (pipe(fd2) == -1)
{
fprintf(stderr, "Problem at pipe: %s\n", strerror(errno));
exit(1);
}
/*create fork*/
pid = fork();
if (pid == -1)
{
fprintf(stderr, "Problem at fork: %s\n", strerror(errno));
exit(1);
}
else if (pid == 0) /*in child*/
{
close(fd2[0]);
close(1);
dup2(fd2[1], 1);
close(fd2[1]);
execl("sample.bin", "sample.bin", NULL);
fprintf(stderr, "Problem at exec: %s", strerror(errno));
exit(1);
}
/* in parent */
close(fd2[1]);
char line[255];
if (n = read(fd2[0], line, 254))
{
printf("%d The message is: %s\n", i, line);
}
close(fd2[0]);
wait(&status);
i++;
}
return 0;
}

Answer 1

I corrected the code, now the output is what I've expected. And of course another problem was that I used at read the same variable n.
I modified from this:
if (n = read(fd2[0], line, 254))
{
printf("%d The message is: %s\n", i, line);
}
To this:
int m;
while((m = read(fd2[0], line, 254) > 0)
{
printf("%d The message is: %s\n", i, line);
}