Linux C program emulating ls command

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I'm trying to create a program that functions similarly to the default ls command using C. I'm having some troubles with one of the attributes (-l) and the formatting currently. For the -l, it is not running the line that is supposed to make it a simple list with the inodes and names (argv[2]) and instead functions the same as -f.

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

struct direct **dirs; 

int main (int argc, char* argv[])
{
  struct stat statb;
  int nent, i;
  int x = 0;
  int y;

  char *F = argv[1];
  char *l = argv[2];
  char *others = argv[3];

  nent = scandir (".", &dirs, NULL, NULL);

  for (i = 0; i < nent; i ++) {       
    if(argv[1]) {   
      stat(dirs[i]->d_name, &statb);  

      if (y == 0) { 
        printf("\n%-16s",dirs[i]->d_name);   
        switch (statb.st_mode & S_IFMT) { 
        case S_IFDIR :                          
          printf("/\t    ");                          
          break;     

        case S_IFREG :                          
          printf("*\t    ");                          
          break;     

        }           
      }                 
      else {  
        printf ("%-16s",dirs[i]->d_name);                   
        switch (statb.st_mode & S_IFMT) { 
        case S_IFDIR :                  
          printf("/\t    ");          
          break;             

        case S_IFREG :                      
          printf("*\t    ");          
          break;      

        }           
      }      

      x++;           
      y = x%5;         
    }         
    else if (argv[2]) {
      printf("I-node = %d   name = %s\n",    dirs[i]->d_ino, dirs[i]->d_name);
}        
else { 

  if (y == 0) {                       
    printf("\n%-16s\t    ",dirs[i]->d_name); 
  }         
  else {       
    printf("%-16s\t    ",dirs[i]->d_name); 
      }    

      x++;    
      y = x%5;      

    }
  } 

  printf("\n");
}

Possible duplicate of http://stackoverflow.com/q/42942357/499581 — l'L'l, Mar 22 '17 at 04:46
Look at `man 3 printf` and particularly the **Format of the format string** section describing the *optional minimum field width*. That allows you to tabulate columns precisely. — David C. Rankin, Mar 22 '17 at 04:46
Thanks!, columns fixed so it looks neat now. Just the -l problem now — nidh0gg, Mar 22 '17 at 04:48
Note that C programs are _not_ scripts (nor are they 'codes', though that isn't a problem here); the `ls` command is not a script, either. You should avoid misusing the names. — Jonathan Leffler, Mar 22 '17 at 05:05
Also note that on Linux, it is unconditionally **wrong** to try using `void main()` — [the correct return type for `main()` in C and C++ is `int`](http://stackoverflow.com/questions/204476/). — Jonathan Leffler, Mar 22 '17 at 05:09

score 0 · Answer 1 · answered Mar 23 '17 at 22:25

0

You will never enter the else clause. If argv[1] is NULL, then argv[2] will be NULL. You just need to parse the arguments.

answered Mar 23 '17 at 22:25

William Pursell

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Linux C program emulating ls command

Resolved

1 Answers1