1. printf's are buffered, try giving a fflush(stdout) after the printf
2. 2nd printf will not print for obvious reason i.e. execl will replace the current process image with a new one.
// Without \n, this goes into a buffer and isn't printed yet.
printf("demo of fork");
// This creates a precise duplicate of the process -- including buffers.
// "demo of...
It is because printf is "buffered," meaning printf will group the output of a process together. While buffering the output for the parent process, the child may also use printf to print out some...