Re: execv fails with EFAULT

["Steve Graegert" <graegerts@gmail.com>]

On 3/7/07, Prasanta Sadhukhan <Prasanta.Sadhukhan@sun.com> wrote:
> Steve Graegert wrote:
>
> > On 3/7/07, Prasanta Sadhukhan <Prasanta.Sadhukhan@sun.com> wrote:
> >
> >> Hi,
> >>
> >> When I tried the following program
> >> void test4()
> >> {
> >>     printf("doing execv\n");
> >>     if(execv("/bin/ls", "-l") == -1)
> >>         printf("exec failed with errno %d\n", errno);
> >>     else
> >>         printf("exec succeed\n");
> >> }
> >
> >
> > Prasanta,
> >
> > A NULL terminated array of arguments must be passed to execv(2) as the
> > prototype indicates: int execv(const char *path, char *const argv[]);
> >
> > For example:
> >
> > #include <unistd.h>
> >
> > char *cmd[] = { "ls", "-l", NULL };
> > int result = execv ("/bin/ls", cmd);
> >
> > An error of EFAULT usually means that an argument points to an illegal
> > address.
> >
> >     \Steve
>
> Thanks Steve... If I have a string like this sprintf(str,
> "-Xparameter:%d %s, value, command"), how to make it NULL terminated. Is
> this declaration char *cmd[] = {str, NULL} and invocation
> execv(path, cmd) correct?

If you mean

        sprintf(str, "-Xparameter:%d %s", value, command);

your declaration and invocation of

        char *cmd[] = {str, NULL}
        execv(path, cmd);

should be OK.

	\Steve

--

Steve Grägert <steve@graegert.com>
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Internet  http://eth0.graegert.com, http://blog.graegert.com
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