Question

The following code is meant to capture the output of an executable in a file and then read in and print out the first 100 bytes of that file. Select all of the bugs in the code. Assume that all system calls succeed and that sizeof(char)-1. pid_t pid = fork(); if (pid == 0) { int in-open("file.txt", READONLY) dup2(STDOUT FILENO, in); execl("./something", "something", "hello!"); l else char buffer[100] int in = open("file.txt'. READ-ONLY): read(buffer, in, 100); The parent does not wait for the child to finish. Two processes cannot have read permissions to a file simultaneously. The order of the arguments to read) is incorrect. read requires a buffer of type 'byte. The file descriptor 'in' is declared twice.

Answer 1

1. The parent does not wait will be taken care by the wait call

pid_t pid = fork();
int in // Removes the redeclarion of in

if (pid == 0){
   in = open("file.txt",READ_ONLY);
   dup2(STDOUT_FILENO,in);
   execl("./something", "something","hello!");
}
else {
      char buffer[100]
      wait(0);    // This will also stop the simultaneous access to the file
      in = open("file.txt",READ_ONLY);
      read(buffer,in,100)
}