execve alwayse produces EFAULT

MD3 · 06-15-2005, 05:30 AM

hi!
i wrote this program: http://lagnusi.net/pub/rungid.c

when i run it, the function

Code:

execve(argv[1], execve_process_argsv, execve_process_envp);

always returns -1 (error) end sets errno to EFAULT, as i can see through

Code:

case EFAULT:
            perror("The filename provided points outside your accesible address space\n");
            break;

why? and how can i avoid this?

thanks

MD3 · 06-15-2005, 09:33 AM

resolved this! there were an argument missing! the function wants the second parameter to be an array of strings where the first is the program name and the last is (char *)0

now... there's the last little thing to examine:
*the program launched is executed 2 times*
as if it was wxecuted normally by the forked process but also by the original process!
someone can explain me why?

itsme86 · 06-15-2005, 10:15 AM

Code:

   my_child_pid = fork();

   if(my_child_pid == -1) {
      perror("Unable to allocate enough memory to fork\n");
      _exit(errno);
   }

   execve_return = execve(argv[1], execve_process_argsv, execve_process_envp);

Because you're calling execve() in the parent and the child. You want to do something like: if(!child_pid) execve(...)

MD3 · 06-15-2005, 01:04 PM

Quote:

Originally posted by itsme86

Code:

my_child_pid = fork(); if(my_child_pid == -1) { perror("Unable to allocate enough memory to fork\n"); _exit(errno); } execve_return = execve(argv[1], execve_process_argsv, execve_process_envp);

Because you're calling execve() in the parent and the child. You want to do something like: if(!child_pid) execve(...)

you're great.. now it works perfectly! thanks