Using Execvp to ls from argument

Question

I'm new to C and trying to understand why this code just return current directory file listing rather than the directory i pass to.

I used c11 standard gcc compiler

and just using ./a.out filedirectory .....

int main(int argc, char *argv[])
{

        char*args[3];
        if(argc > 1)
        {
                args[0] = "/bin/ls";
                args[1] = "-l";
                if(argv[2])
                {
                        args[2] = argv[2];
                        args[3] = NULL;
                }
                else
                {
                        args[2] = NULL;
                }

                execvp(args[0],args);

                return 0;
        }
}

Did you *carefully* read the documentation of [execvp(3)](http://man7.org/linux/man-pages/man3/execvp.3.html) and of [execve(2)](http://man7.org/linux/man-pages/man2/execve.2.html)? BTW, your question lacks a `Linux` or `Posix` tag since `execvp` is not defined in the C11 standard. Read also [*Advanced Linux Programming*](http://advancedlinuxprogramming.com/) — Basile Starynkevitch, Jun 03 '17 at 05:21
As usual, compile with all warnings & debug info (`gcc -Wall -Wextra -g`) and **use the debugger `gdb`** (so spend several hours learning it). You may also find [strace(1)](http://man7.org/linux/man-pages/man1/strace.1.html) useful. — Basile Starynkevitch, Jun 03 '17 at 05:22
Side note: with `char *args[3]`, accessing `args[3]` is undefined. — dhke, Jun 03 '17 at 05:40
Thank you Basile, I'll check the documentation of execvp again. — Shawn Chen, Jun 03 '17 at 07:00

Basile Starynkevitch · Answer 1 · 2017-06-03T08:23:40.000

You did not tell how are your running your program myprog. Let's suppose you run it as (for example) ./myprog /tmp/.

In the C programming language (read n1570) array indexes start at 0!

Then when entering your main you have

argc is 2
argv[0] is the "./myprog" string
argv[1] is the "/tmp/" string
argv[2] is the NULL pointer (this is guaranteed by the C specification of main and implemented by crt0) since it is the main function

^{(In all other cases in C, arrays of pointers are not implicitly NULL terminated, but main's argv is special)}

Follow with a pencil or with the gdb debugger your program source. The first if (argc > 0) succeeds but the second one if (argv[2]) fails. Then you have args[0] set to "/bin/ls", args[1] set to "-l" and args[2] set to NULL before the execvp call.

See execvp(3), execve(2) and read Advanced Linux Programming

You really should take the habit of compiling with all warnings & debug info (gcc -Wall -Wextra -g with GCC, e.g. compile yoursource.c source file into myprog executable with gcc -Wall -Wextra -g yoursource.c -o myprog) and learn how to use the debugger gdb.

BTW, your use of args[3] is undefined behavior (out of bound indexing or buffer overflow), so you should be scared, since args was declared as an array of dimension 3 (so valid indexes are only 0, 1 and 2). In your case a recent GCC would warn you.

There is also the strange bit of logic of the nested `if..else` inside of `if (argc > 0)` (which will always be true). Instead, a single `if (argc > 1) ... else ...` could accomplish what looks to be his goal. — David C. Rankin, Jun 03 '17 at 06:08

Using Execvp to ls from argument

1 Answers1