execve() different program output printing

Question

I've recently been working with the execve() system call and I've been trying to figure out why it would print the full path of the function I called.

For example, when calling ls with execve :

/usr/bin/ls: cannot access 'nofile': No such file or directory

I mean, if I think about it, it's quite logical. If I call /usr/bin/ls as filename in execve , this is what the pathname actually is.

But why is the ls system call output like this:

ls: cannot access[...]

and not like mine?

I was wondering if there was any solution to find out the "name" of the program I just called, not to print the full path of the program.

As it's execve 's choice and not mine, at least I guess it is, I'm not sure about it.

Thanks.

Answer 1

I used of execvp function in the following code and built it:

#include  <stdio.h>
#include  <sys/types.h>

void  parse(char *line, char **argv)
{
     while (*line != '\0') {       /* if not the end of line ....... */ 
          while (*line == ' ' || *line == '\t' || *line == '\n')
               *line++ = '\0';     /* replace white spaces with 0    */
          *argv++ = line;          /* save the argument position     */
          while (*line != '\0' && *line != ' ' && 
                 *line != '\t' && *line != '\n') 
               line++;             /* skip the argument until ...    */
     }
     *argv = '\0';                 /* mark the end of argument list  */
}

void  execute(char **argv)
{
     pid_t  pid;
     int    status;

     if ((pid = fork()) < 0) {     /* fork a child process           */
          printf("*** ERROR: forking child process failed\n");
          exit(1);
     }
     else if (pid == 0) {          /* for the child process:         */
          if (execvp(*argv, argv) < 0) {     /* execute the command  */
               printf("*** ERROR: exec failed\n");
               exit(1);
          }
     }
     else {                                  /* for the parent:      */
          while (wait(&status) != pid)       /* wait for completion  */
               ;
     }
}

void  main(void)
{
     char  line[1024];             /* the input line                 */
     char  *argv[64];              /* the command line argument      */

     while (1) {                   /* repeat until done ....         */
          printf("Shell -> ");     /*   display a prompt             */
          gets(line);              /*   read in the command line     */
          printf("\n");
          parse(line, argv);       /*   parse the line               */
          if (strcmp(argv[0], "exit") == 0)  /* is it an "exit"?     */
               exit(0);            /*   exit if it is                */
          execute(argv);           /* otherwise, execute the command */
     }
}

Now, when i run the program in terminal , the result is something like this (I tried with ls command):

msi@ubuntu:~/Desktop$ ./output
Shell -> ls

1.txt
2.jpg
Qt-Creator.run
.
.
.

It also does worked with this parameter (Full path of ls program):

Shell -> /bin/ls
1.txt
2.jpg
Qt-Creator.run
.
.
.

Answer 2

I found out the issue, but as It's in an answer to BattleTested - закалённый в бо , I figured that answering it here would be better for anyone having this issue.

Basically, to call execve , I needed a pathname (here /usr/bin/ls ), an array of arguments (which contains the program name as first argument, not just the arguments or a NULL array if none) and the environment.

Here for ls with the -l should've been:

pathname: "/usr/bin/ls"

arguments: {"/usr/bin/ls", "-l", NULL};

and your environment (NULL terminated).

The thing is, execve() will use the first argument of your arguments array , which here is /usr/bin/ls . Simply modifying it to the program's name, and not path, fixed it.