Traverse a directory tree and output a list of all files

Question

Write the shortest code that traverses a directory tree and outputs a flat list of all files.

• It should traverse the whole directory tree
• The order in which it enters sub-directories doesn't matter
• The directories should not be output
• The output should be a plain list — the definition is flexible, but it shouldn't contain irrelevant data
• You are free to use any programming technique

Answer 1

find 14 characters (12 on GNU)

find . -type f

Run this in the directory you want to list. Add -printf '%f\n' to only print names without the path.

Non-portable: find -type f.

Answer 2

Pure Bash, 7

"ls -R" and "find" are the obvious, already mentioned common shell utilities. For completeness, we also have this pure bash option (no utility programs):

echo **

This requires the "globstar" shell option to be set with shopt -s globstar.

Note:

This works only for Bash 4.0 and greater.

Answer 3

Python

Aleksi Torhamo's answer is the correct and idiomatic one. My answer is just an exercise of what if there was no os.walk.

I was inspired by an elegant recursive function for Emacs Lisp that does the same.

from os import *
from os.path import *
def walk(path):
    return [basename(f) for f in listdir2(path) if not isdir(f)] +\
        sum([walk(f) for f in listdir2(path) if isdir(f)], [])

listdir2 here is a non-existent function that returns absolute paths of files in a directory. Unfortunately, listdir doesn't return absolute paths, therefore i had to manually enter and exit directories with chdir, so my result wasn't a slick one-liner.

from os import *
from os.path import *
def walk(path, old_path):
    chdir(path)
    result = [f for f in listdir(path) if not isdir(f)] +\
        sum([walk(join(path, f), getcwd()) for f in listdir(path) if isdir(f)], [])
    chdir(old_path)
    return result

To understand it you must read about os.chdir, os.getcwd, os.listdir, os.path.isdir, os.path.basename, os.path.join, sum. The function creates a list of files and recursively repeats this for sub-directories. The list of sub-directories is summed up and appended to the list of original files.

We iterate thru list 2 times, this is inefficient. Better to partition the list into files and directories in one go. This answer proposes a smart use of reduce:

reduce(lambda x, y: x[not p(y)].append(y) or x, l, ([], []))

Attempt #1 (179 171)

from os import*
def w(p,o):chdir(p);r=[f for f in listdir(p)if not path.isdir(f)]+sum([w(path.join(p,f),getcwd())for f in listdir(p)if path.isdir(f)],[]);chdir(o);return r

Attempt #2 (190 178)

from os import *
def u(p,o):chdir(p);a,b=reduce(lambda x,y:x[path.isdir(y)].append(y)or x,listdir(p),([],[]));r=a+sum((u(path.join(p,c),getcwd())for c in b),[]);chdir(o);return r

Answer 4

DOS (11 characters)

dir/s/b/a-d

I love the simplicity, /s allows dir to traverse through all subfolders, while /b suppresses the header and directory information. /a-d suppresses outputting of directories. A quick check to see if the additional space was required, and I was happy enough to submit this...

Answer 5

Python3 - 53 characters

Two equally long ways for doing it:

import os;print(*sum((x[2]for x in os.walk('.')),[]))

and

import os;print(*sum(list(zip(*os.walk('.')))[2],[]))

These print the filenames separated by spaces on one line, I assumed that counts as a "flat list".

Answer 6

C - 203 194

This example works when compiled with TCC:

t(char*p){DIR*d;struct dirent*e;if(d=opendir(p))while(e=readdir(d)){char*n=e->d_name,f[PATH_MAX];sprintf(f,"%s/%s",p,n);if((*n++-46|*n&(*n++-46|*n))&&!t(f))printf("%s\n",f);}return!closedir(d);}

This recursive function displays the path of every file present in the folder which path has been given as a parameter.

Note: one character had to be added to get the same result with GCC:

int t(char*p){DIR*d;struct dirent*e;if(d=opendir(p))while(e=readdir(d)){char f[PATH_MAX];char*n=e->d_name;sprintf(f,"%s/%s",p,n);if((*n++-46|*n&&(*n++-46|*n))&&!t(f))printf("%s\n",f);}return!closedir(d);}

I replaced & with && in the condition. I guess the & operator has priority over && in TCC syntax, while the opposite occurs with GCC.

See the full working example here: http://ideone.com/lGBtJw

C - 220 215 (full program with include)

I liked that Coaumdio used a full program to answer... so I started adapting mine, here is the result:

#include<dirent.h>
main(i,char**p){DIR*d=opendir(*++p);char*n,f[4096],*a=f;struct dirent*e;d||puts(*p);if(d)while(e=readdir(d))sprintf(f,"%s/%s",*p,n=e->d_name),(*n++-46|*n&&(*n++-46|*n))&&main(i,&a-1);closedir(d);}

Expanded version:

#include<dirent.h>

int main(int i,char**p){
    DIR*d=opendir(*++p);
    char f[4096],*n,*a=f;
    struct dirent*e;
    d||puts(*p);
    if(d)while(e=readdir(d))
        sprintf(f,"%s/%s",*p,n=e->d_name),
        (*n++-46|*n&&(*n++-46|*n))&&main(i,&a-1);
    closedir(d);
}

Usage:

gcc list_files.c -o list_files
./list_files /var

Answer 7

Bash, 18

ls -R

Windows CMD.

dir /S

Compliant version

ls -R1|grep -v ^./

Answer 8

C, 216 228 chars (#include included)

The other C solution contians nice tricks, but as I like compilable complete C solutions, here is my shot:

#include <dirent.h>
main(int _,char**v){char n[4096],*a=n,*b;DIR*d;struct dirent*e;(d=opendir(*++v))||puts(*v);while(d&&(e=readdir(d)))b=e->d_name,sprintf(n,"%s/%s",*v,b),strcmp(".",b)&&strcmp("..",b)&&main(0,&a-1);}

It's also a recursive traversal algorithm, and it's using the main function as recursive function. The main function tries to open the path contained in argv[1]. If it fails, it prints it, otherwise, it calls main on every sub-element (if it's different from "." or "..").

Usage

It's a standalone, working C program:

$ gcc dir.c -o dir
$ ./dir /var.log   
/var/log/kern.log.1
/var/log/alternatives.log.1
/var/log/user.log.2.gz
/var/log/nvidia-installer.log
/var/log/debug.4.gz
/var/log/wtmp.1
/var/log/lastlog
/var/log/messages.2.gz
/var/log/vmware-installer
/var/log/pm-powersave.log
/var/log/syslog.4.gz
/var/log/news/news.err
/var/log/news/news.notice
/var/log/news/news.crit
/var/log/apt/history.log.3.gz
/var/log/apt/history.log.9.gz
/var/log/apt/history.log.4.gz
/var/log/apt/term.log.10.gz
/var/log/apt/history.log.10.gz
/var/log/apt/term.log.4.gz

Tips

I used a few (somewhat classic) C tricks:

• replace if(condition)statement with contition && statement
• replace if(!condition)statement with condition || statement
• use the , operator to save { and } in the while code

Here is a readable version:

#include <dirent.h>
main(int _,char**v) {
    char n[4096], *a=n,*b;
    DIR *d; 
    struct dirent *e; 
    (d = opendir(*++v)) || puts(*v);
    while(d && (e = readdir(d)))
        b = e->d_name,
        sprintf(n, "%s/%s", *v, b), 
        strcmp(".", b)&&strcmp("..", b)&&
            main(0, &a-1);
}

Edit

I have to insert a closedir(d); statement at the end, because a limit is reached when too much folders are opened (wild guess: the open files limit) and the the program stops. This happens only for folder containing many other folders.

the code is now:

#include <dirent.h>
main(int _,char**v){char n[4096],*a=n,*b;DIR*d;struct dirent*e;(d=opendir(*++v))||puts(*v);while(d&&(e=readdir(d)))b=e->d_name,sprintf(n,"%s/%s",*v,b),strcmp(".",b)&&strcmp("..",b)&&main(0,&a-1);closedir(d);}

Answer 9

Ruby 18 characters

puts Dir['./**/*']

Answer 10

Groovy 58

new File("/").eachFileRecurse(){if(it.isFile())println it}

Answer 11

CMD - 32 Bytes

Another CMD / DOS method -

for /r %a in (*)do @echo %~dpnxa

The /r switch recursively looks through a directory (defaulting to current directory), while %~dpnxa outputs %a, the current directory or file in the loop, to the following output format -

d - drive
p - path
n - name
x - extension

Because x is specified it will only output files, ignoring directories.

Answer 12

Groovy 45

Now even shorter:

new File(".").traverse{if(it.file)println it}

Uses one of the traverse() methods available since Groovy 1.7.1.

Groovy 46 ;-)

Another nice solution, if you don't count the required import statement, is this one:

import groovy.io.FileType.*
new File(".").traverse type:FILES,{println it}

Answer 13

SmileBASIC, 78 bytes

DIM F$[0],D$[0]FILES"//",D$@L
FILES POP(D$),F$WHILE LEN(F$)?POP(F$)WEND
GOTO@L

Displays a * or (please don't remove my spaces) before each filename, depending on whether it's a text file (is this allowed?)

Answer 14

JAVA 87

void t(File f){if(f.isFile())System.out.println(f);else for(File c:f.listFiles())t(c);}