I am trying to implement a simple shell program that runs user input commands. I want the user to enter "ls" or "dir" and have the shell run /bin/ls or /bin/dir . For the execve argument what would be correct:
char *args[] ={"/bin/", "ls", NULL};
//Fork, pid, etc...then
execve(args[0], args+1, NULL);
Or would it be something different?? I've see some people using /bin/ls as the pathname and then no arguments or ls as the pathname and \bin for the environment? I tried what I had above and it didn't work so I want to know if it is the arguments I am sending it or whether I should be looking elsewhere in my code for the problem. Note I am not interested in other variations of execve such as execvp. I am on a Linux system. Thanks
PATHNAME
The pathname in execve() must be the full path to the executable, such as /bin/ls. If using execvpe(), you could use ls alone as the pathname, but as you already specified, you don’t want to use that.
ARGUMENTS
The arguments should be an array of strings, one for each space-separated argument specified on the command line. The last one should be NULL. The first argument should be the pathname itself. For example:
char* args[] = {"/bin/ls", "-la", "foo/bar", NULL};
ENVIRONMENT
The environment variables cannot be omitted when using execve(). In some implementations, NULL can be passed as the last argument to execve(), but this is not standard. Instead, you should pass a pointer to a null pointer; essentially an empty array of environment variables.
Putting it together
char *args[] ={"/bin/ls", "-foo", "bar", NULL};
//Fork, pid, etc...then
char* nullstr = NULL;
execve(args[0], args, &nullstr);
From execve [emphasis added]:
int execve(const char *pathname, char *const argv[],
char *const envp[]);
execve() executes the program referred to by pathname. This
causes the program that is currently being run by the calling
process to be replaced with a new program, with newly initialized
stack, heap, and (initialized and uninitialized) data segments.
pathname must be either a binary executable, or a script starting
with a line of the form:
#!interpreter [optional-arg]
For details of the latter case, see "Interpreter scripts" below.
argv is an array of pointers to strings passed to the new program
as its command-line arguments. By convention, the first of these
strings (i.e., argv[0]) should contain the filename associated
with the file being executed. The argv array must be terminated
by a NULL pointer. (Thus, in the new program, argv[argc] will be
NULL.)
In your case, the pathname should be "/bin/ls" and not "/bin/". If you want to pass any command line argument with the command, you can provide first argument them with argv vector index 1, second argument with index 2 and so on and terminate the argument vector with NULL.
A sample program which replace the current executable image with /bin/ls program and runs /bin/ls testfile:
#include <stdio.h>
#include <unistd.h>
int main (void) {
char *args[] = {"/bin/ls", "testfile", NULL};
// here you can call fork and then call execve in child process
execve(args[0], args, NULL);
return 0;
}
Output:
# ./a.out
testfile
Related
I'm new to using linux and the bash command line but I was trying to see whether I could execute basic bash commands (or program) from a c file. My code is really simple and I could've sworn that it was working but suddenly it is not acknowledging that the path even exists.
char* arg[] = {"ls", "-l"};
char* environ[] = {"PATH=./", (char*)0};
execve("/bin/ls", arg, environ);/*execute the command 'ls -l' in curdir*/
I've tried to set as PATH=/bin but it's not even executing the command in that directory. it just returns a similar error.
NOTE: I have tried
char* environ[] = {"PATH=./", NULL};
I've even tried using the envp from main() and that STILL doesn't work.
This error message ...
ls: cannot access 'PATH=./': No such file or directory
... indicates that the ls utility is attempting to access a file named "PATH=./", which it does not find. That is a manifestation of the undefined behavior arising from your code ...
char* arg[] = {"ls", "-l"};
char* environ[] = {"PATH=./", (char*)0};
execve("/bin/ls", arg, environ);/*execute the command 'ls -l' in curdir*/
... on account of execve() expecting and relying upon the argument list to which arg points being terminated by a null pointer.
Although it is somewhat fraught to try to rationalize or interpret undefined behavior, you can imagine that the contents of arrays arg and environ are laid out one immediately after the other in memory, so that the combined representation of these two is the same as the representation of a four-element array of char *. This view is perhaps useful for understanding why the arg and env arrays must be terminated by null pointers in the first place.
The fix is to append a null pointer to the value of arg:
char* arg[] = {"ls", "-l", NULL};
Note also that in this particular case there is no apparent advantage to specifying an environment, so you could simplify by using execv() instead.
Note, too, that path elements other than / itself should not contain a trailing / character. This is not part of a correct name for a directory, and although you can often get away with it, it is poor form.
Put NULL in the end of arg array.
char* arg[] = {"ls", "-l", NULL};
The NULL is used to mark the end of the array.
Please always post a complete program.
As said in the man page, and as said in the comments above, you need a NULL at the end of the arguments list.
You do not need to pass anything in envp[] since you are running just ls
By convention you should pass the full path of the executable in argv[0]
In short, this works
#include <stdio.h>
#include <unistd.h>
int main(void)
{
char* arg[] = { "/bin/ls", "-l", NULL };
execve("/bin/ls", arg, NULL);
return 0;
}
I have a program written by my professor that prints the working directory (pwd) by using execve(), but I don't understand the parameters.
pid_t pid = fork();
if(pid <0)
perror(NULL);
else if(pid == 0)
{
char*argv[] = {"pwd",NULL};
execve("/bin/pwd",argv,NULL);
perror(NULL);
}
else
printf("Im the parent!");
return 0;
}
"/bin/pwd" gives the path to the executable that will be executed.
This means that it will call the pwd function, doesn't it?
Then why do I need to have the parameter pwd?
Couldn't the program run without that parameter?
By convention, the first argument passed to a program is the file name of the executable. However, it doesn't necessarily have to be.
As an example, take the following program:
#include <stdio.h>
int main(int argc, char *argv[])
{
int i;
printf("number of arguments: %d\n", argc);
printf("program name: %s\n", argv[0]);
for (i=1; i<argc; i++) {
printf("arg %d: %s\n", argv[i]);
}
return 0;
}
If you run this program from another like this:
char*argv[] = {"myprog", "A", "B", NULL};
execve("/home/dbush/myprog",argv,NULL);
The above will output:
number of arguments: 3
program name: myprog
arg 1: A
arg 2: B
But you could also run it like this
char*argv[] = {"myotherprog", "A", "B", NULL};
execve("/home/dbush/myprog",argv,NULL);
And it will output:
number of arguments: 3
program name: myotherprog
arg 1: A
arg 2: B
You can use the value of argv[0] as a way to know how your program was called and perhaps expose different functionality based on that.
The popular busybox tool does just this. A single executable is linked with different file names. Depending on which link a user used to run the executable, it can read argv[0] to know whether it was called as ls, ps, pwd, etc.
The execve man page has some mention of this. The emphasis is mine.
By convention, the first of these strings should contain the filename associated with the file being executed.
That is, it is not a actually mandatory for the first argv to be the filename. In fact one can test that by changing the argv[0] to any string in the example code and the result will still be correct.
So it really is just a convention. Many programs will use argv[0] and expect it to be the filename. But many programs also do not care about argv[0] (like pwd). So whether argv[0] actually needs to be set to the filename depends on what program is being executed. Having said that, it would be wise to always follow the convention to play nicely with almost everyone's long held expectations.
From execve man page: http://man7.org/linux/man-pages/man2/execve.2.html
argv is an array of argument strings passed to the new program. By
convention, the first of these strings (i.e., argv[0]) should contain
the filename associated with the file being executed. envp is an
array of strings, conventionally of the form key=value, which are
passed as environment to the new program. The argv and envp arrays
must each include a null pointer at the end of the array.
So, argv is treated as command line args for new program to execute with.
Since by default, for a linux binary invoked with arguments, these args are accessed through argc/argv, where argv[0] holds the program name.
I think this is to keep the behavior parity to match with default case (prog invoked with arguments).
From the source:
https://git.kernel.org/pub/scm/linux/kernel/git/torvalds/linux.git/tree/fs/exec.c#l1376
The argv passed to execve is used to construct argv for the about to be launched binary.
I need to see a concrete example of how to specify the environment for execve() in a c program. In my class, we are writing a program that will utilize both standard LINUX executables and our own executables. Thus, the environment searching PATH will have to contain tokens for both types of executables. I cannot find a good example of how to specify the environment (third argument) for execve() as every article seems to suggest we use execvp() or *clp() or *cl(), etc., instead.
In my project, we must use execve().
Right now, I'm just trying to get execve() to work for a basic "ls" command so that I can get it to work later for any and all executables.
Here is a snippet of my experiment code:
else if(strcmp(tokens[0], "1") == 0) {
char *args[] = {"ls", "-l", "-a", (char *)0};
char *env_args[] = {"/bin", (char*)0};
execve(args[0], args, env_args);
printf("ERROR\n");
}
Each time command "1" is entered in my shell, I see my error message. I suspect this is because of the way I am declaring env_args[].
Can someone show me a good example of how to implement execve() with a specified command searching environment?
here is the documentation on execve() function http://linux.die.net/man/2/execve
it says:
int execve(const char *filename, char *const argv[], char *const envp[]);
envp is an array of strings, conventionally of the form
key=value, which are passed as environment to the new program.
but in your program env_args does not look like key=value
So probably you should define env_args by the following way:
char *env_args[] = {"PATH=/bin", (char*)0};
or just
char *env_args[] = { (char*)0 };
So I'm writing a program where the arguments are as follows:
program start emacs file.c
or even
program wait
In essence, the first argument (argv[0]) is the program name, followed by user inputs.
Inside my code, I invoke execvp. Thing is, I'm not entirely sure I'm invoking the right arguments.
if (pid == 0) {
execvp(argv[1], argv); //Line of interest
exit(1);
}
are argv[1] and argv the correct arguments to pass for the functionality described above? I looked at the man page and they make sense but might not be correct for this case.
Thank you!
In your main, argv will be like this in the first example:
argv[0] = "program";
argv[1] = "start";
argv[2] = "emacs";
argv[3] = "file.c";
argv[4] = NULL;
In execv you want to execute the program "start" with args "emacs file.c", right?. Then the first parameter should be argv[1] - "start" and the second one an array with this strings: {"start", "emacs", "file.c", NULL}. If you use argv, you include the "program" string in argv[0].
You can create a new array and copy these parameters or use the address of argv[1] like this:
execvp(argv[1], &argv[1]); //Line of interest
The only thing that might be an issue is that argv[0] in argv passed to execvp won't match argv[1] (the first argument). Otherwise, it looks okay.
Imagine calling program cat file.txt. In your program, argv will be {"program", "cat", "file.txt", NULL}. Then, in cat, even though the binary called will be cat, argv will still be {"program", "cat", "file.txt", NULL}.
Since cat tries to open and read each argument as a file, the first file it'll try to open is cat (argv[1]), which isn't the desired behavior.
The simple solution is to use execvp(argv[1], argv+1) - this essentially shifts the argument array to the left by one element.
My understanding is that you want to take a specific action based on the second command-line argument (argv[1]). If the second argument is 'start', your program should start the executable named argv[2] with the arguments provided thereafter (right?). In this case, you should provide execvp with the executable name (argv[2]) [1] and a list of arguments, which by convention starts with the name of the executable (argv[2]).
execvp(argv[2], &argv[2]) would implement what we have described in the last paragraph (assuming this is what you intended to do).
[1] execvp expects 2 arguments as you know. The first is a filename; if the specified filename does not contain a slash character (/), execvp will do a lookup in the PATH environment variable (which contains a list of directories where executable files reside) to find the executable's fully-qualified name. The second argument is a list of command-line arguments that will be available to the program when it starts.
I am trying to use execve to run the ls command. Currently I'm running it with the following arguments:
execve(args[0], args, env_args)
//args looks like {"ls", "-l", "-a", NULL}
//env_args looks like {"PATH=/bin", "USER=me", NULL}
What I expected this to do was run the ls command using my new env_args meaning that it would look up ls in my PATH. However, this code actually doesn't do anything and when I run the code it just returns to my command prompt without output.
Using the same args[] I was using execvp and ls worked and searched my current path.
Can you tell me what I am doing wrong?
What I am trying to do is write my own shell program where I can create and export my own environment and have exec use the environment I have defined in a char**. Essentially I am writing my own functions to operate on env_args to add and remove vars and when I call exec i want to be able to call exec on {"ls", "-l", NULL} and have it look down my new environments path variable for a valid program called ls. I hope this explains what I am doing a little better. I don't think the extern environ var will work for me in this case.
The execve() function does not look at PATH; for that, you need execvp(). Your program was failing to execute ls, and apparently you don't report failures to execute a program after the execve(). Note that members of the exec*() family of functions only return on error.
You'd get the result you expected (more or less) if you ran the program with /bin as your current directory (because ./ls - aka ls - would then exist).
You need to provide the pathname of the executable in the first argument to execve(), after finding it using an appropriate PATH setting.
Or continue to use execvp(), but set the variable environ to your new environment. Note that environ is unique among POSIX global variables in that is it not declared in any header.
extern char **environ;
environ = env_args;
execvp(args[0], &args[0]);
You don't need to save the old value and restore it; you're in the child process and switching its environment won't affect the main program (shell).
This seems to work as I'd expect - and demonstrates that the original code behaves as I'd expect.
#include <stdio.h>
#include <unistd.h>
extern char **environ;
int main(void)
{
char *args[] = { "ls", "-l", "-a", NULL };
char *env_args[] = { "PATH=/bin", "USER=me", NULL };
execve(args[0], args, env_args);
fprintf(stderr, "Oops!\n");
environ = env_args;
execvp(args[0], &args[0]);
fprintf(stderr, "Oops again!\n");
return -1;
}
I get an 'Oops!' followed by the listing of my directory. When I create an executable ls in my current directory:
#!/bin/sh
echo "Haha!"
then I don't get the 'Oops!' and do get the 'Haha!'.