UNIX command prompt

[repairman2003]

I'm writing a command prompt for unix and I've run into some problems:
#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <stdlib.h>
#define EXIT "exit"
#define DELIMITER " "
we are in
int main(argc, argv)
{
//Initilzation of varibles
char **arguments = (char **) malloc (8);
char *command_buffer = (char *) malloc (256);
char *arg_ptr;
int i = 0;
int pid;
int nbytes = 256;
int not_exiting = 1;

do{
//prints prompt and gets line of text
printf("sim> ");
getline(&comman d_buffer, &nbytes, stdin);

//Special workaround
//replaced the '\n' at the end of the input, which was giving errors
//with '\0'
command_buffer[strlen(command_ buffer) -1] = '\0';

//picks line apart and
//replaces spaces with '\0' and adds pointers to argument[]
do{
if(i == 0)
{
arguments[i] = strtok(command_ buffer, DELIMITER);
i++;
}
else
{
arg_ptr = strtok(NULL, DELIMITER);

if(arg_ptr != NULL)
{
arguments[i] = arg_ptr;
i++;
}
else
{
i++;
break;
}
}
}while(1);

//are we exiting?
not_exiting = strcmp(command_ buffer, EXIT);
if(not_exiting)
{
//create child process
pid = fork();

if(pid == 0) //fork() == 0 means child process
{
execvp(*argumen ts, arguments);
exit(0);
}
else if(pid > 0) //fork() > 0 means we are in parent process
{

wait(NULL);
}
else //otherwise we have error
{
fprintf(stderr, "fork() returned error");
exit(1);
}
}
else
{
printf("Exiting \n");
}
}while(not_exit ing);

return 0;
}

The problems
1. The workaround listed above.
2. One argument commands work fine but multiple don't. Eg. ls -a
works fine, ls -al works fine but ls -a -l doesn't work. Any
subsequent commands that don't have arguments don't work either, eg.
ls works, then ls -a works, but ls again doesn't work.

Any help is greatly appriciated!

[Walter Roberson]

In article <11************ **********@g43g 2000cwa.googleg roups.com>,
<re***********@ gmail.com> wrote:

I'm writing a command prompt for unix and I've run into some problems:

Try comp.unix.progr amming . You used several calls which are
not part of standard C (such as fork()), and it is best to take
the questions to people who are familiar with the meaning of those
calls.

comp.lang.c prefers to restrict itself to standard C.
--
Okay, buzzwords only. Two syllables, tops. -- Laurie Anderson

[repairman2003]

Ok thanks, I'll post it up there. But I know the unix calls all work,
just more or less a matter of the reading in from user and handling the
strings, I'm not used to doing that in C.

[Mark McIntyre]

On 28 Jan 2006 21:23:42 -0800, in comp.lang.c ,
re***********@g mail.com wrote:

Ok thanks, I'll post it up there. But I know the unix calls all work,
just more or less a matter of the reading in from user and handling the
strings, I'm not used to doing that in C.

I suggest you try a simpler example without all the unix-ey bits, just
reading user input and handling some strings. If you get stuck with
that, then post here.
Mark McIntyre
--
"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan

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[serrand]

re***********@g mail.com wrote:

I'm writing a command prompt for unix and I've run into some problems:


You're doing as if array of char * was char** (pointer on char *)
see faq 6.19 (thanks Flash Gordon)
//Initilzation of varibles /* instead of */ char **arguments = (char **) malloc (8); /*use this */
char * arguments[8] = {NULL};
char *command_buffer = (char *) malloc (256);
char *arg_ptr;
int i = 0;
int pid;
int nbytes = 256;
int not_exiting = 1;

do{
//prints prompt and gets line of text
printf("sim> ");
getline(&comman d_buffer, &nbytes, stdin);

//Special workaround
//replaced the '\n' at the end of the input, which was giving errors
//with '\0'
command_buffer[strlen(command_ buffer) -1] = '\0';

//picks line apart and
//replaces spaces with '\0' and adds pointers to argument[]
do{
if(i == 0)
{
arguments[i] = strtok(command_ buffer, DELIMITER);
i++;
}
else
{
arg_ptr = strtok(NULL, DELIMITER);

if(arg_ptr != NULL)
{
arguments[i] = arg_ptr;
i++;
}
else
{
i++;
break;
}
}
}while(1);

//are we exiting?
not_exiting = strcmp(command_ buffer, EXIT);
if(not_exiting)
{ //create child process
pid = fork();

if(pid == 0) //fork() == 0 means child process
{
execvp(*argumen ts, arguments);
exit(0);
}
else if(pid > 0) //fork() > 0 means we are in parent process
{

wait(NULL); while (i) arguments[i--] = NULL; }
else //otherwise we have error
{
fprintf(stderr, "fork() returned error");
exit(1);
}
}
else
{
printf("Exiting \n");
}
}while(not_exit ing);

return 0;
}

The problems
1. The workaround listed above.
2. One argument commands work fine but multiple don't. Eg. ls -a
works fine, ls -al works fine but ls -a -l doesn't work. Any
subsequent commands that don't have arguments don't work either, eg.
ls works, then ls -a works, but ls again doesn't work.

Any help is greatly appriciated!

Fine... but we can't see the command's result...
But... it's not the place here to tell you more on this topic ;-)
Xavier

[Arndt Jonasson]

re***********@g mail.com writes:

I'm writing a command prompt for unix and I've run into some problems:

Three suggestions:
1) Use a debugger;
2) Add debug print statements at appropriate places;
3) Write a small C (or shell or Perl or whatever) program which prints
out its arguments in a way that you know exactly what it received,
then call it in various ways from your shell ("command prompt" to me
sounds like just, well, the prompt).