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Doing an ls in C

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Jayjaynam Posted messages 2 Status Member -  
Jayjaynam Posted messages 2 Status Member -
Hello, I'm trying to replicate the functionality of ls in C, but I'm facing an issue. Here is my code: 


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

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

 DIR* dir = NULL;
 struct dirent* fileRead = NULL;
 if((dir = opendir("/home/jayjay/projetres")) == NULL){
 perror("The file could not be opened");
 exit(1);
 }

 while((fileRead = readdir(dir)) != NULL){
 if(strcmp(fileRead->d_name, ".") != 0 && strcmp(fileRead->d_name, "..") != 0 && fileRead->d_name[0] != "."){
 printf("%s\n", fileRead->d_name);
 }
 }
 closedir(dir);
}

The problem is in the line where I compare strings. The 3rd comparison is wrong, and I know it, but actually, I just want to compare my string to .* to remove all hidden files from the list that will be displayed like ls does. This is the point where I would like your help because I have no idea how to do that.

Thanks in advance for your answers.
Jayjaynam

2 answers

  1.
    [Dal] Posted messages 6122 Registration date   Status Contributor Last intervention   1 108
     
    Hi Jayjaynam,

    https://man7.org/linux/man-pages/man3/readdir.3.html

    Since the 
    dirent
    structure contains 
    d_name[256]
    , which is a C string, you just need to check that the first character of the d_name array is not a dot.

    You can do this simply like this: 

    if ( (strcmp(fichierLu->d_name, "..") != 0) && (fichierLu->d_name[0] != '.') ) {


    This also allows you to remove the test 
    (strcmp(fichierLu->d_name, ".") != 0)
    since this case is anyway covered by the inequality test of the first character with the dot.

    Dal
    2
  2. Jayjaynam Posted messages 2 Status Member 1
     
    Oh yes, indeed, I completely forgot the basics of the basics... Thanks to you, it's working well and I can continue practicing.
    The subject can therefore be closed.

    Thank you very much and have a great day.
    Jayjaynam.
    1