Shortest code to dump a file into workable memory

Question

This is a slightly different code golf from the usual problems. I'm forced to work with Java at my workplace and I'm increasingly annoyed at what I must do to read a file into memory to a String so I can operate on it in a normal way.

The golf is this: What is the shortest possible function to dump a file into your language's smallest data structure of characters?

Java would be char[], C++ would be char*, Haskell would be a [Char], etc. I don't care about complete programs or main() functions, but you do have to open and close the file.

inb4 Unix's >

Answer 1

Python

a=open('filename.txt').read()

The garbage collector will close the file.

Answer 2

C#

var a = System.IO.File.ReadAllText("filename.txt");

a is a string that will contain the contents of filename.txt

Answer 3

PHP

$a=file('foo.txt');

$a now contains an array of strings of the lines of foo.txt.

Similarly:

$a=file_get_contents('foo.txt');

$a now contains a single string of the contents of foo.txt.

Answer 4

C (for unixoid systems)

#include <stdlib.h>
#include <sys/mman.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <fcntl.h>

/* inside a function */
struct stat fd_stat;
int fd;
void *buffer;

int fd = open("my.file",O_RDONLY);
if (fd<0) exit(EXIT_FAILURE);
fstat(fd,fd_stat);
buffer = mmap(NULL,fd_stat->st_size,PROT_READ,MAP_SHARED,fd,0);
if (buffer==NULL) exit(EXIT_FAILURE);

This ridiculously long piece of code creates a buffer buffer of length fd_stat->st_size that contains the file. You probably won't use C for code-golfing anyway..

Answer 5

J

a=.1!:1<'filename.txt'

a now contains a string representing the contents of filename.txt.

Answer 6

Scala

In Scala, you can get a iterator easily:

scala> io.Source.fromFile("Cg6367.scala").getLines 
res274: Iterator[String] = non-empty iterator

then you can iterate over a collection of lines with a for loop, or use one of the other 200 methods, usable on iterators, like map, forall, filter, first, head, last, drop, groupBy, sortBy, ... - you name it. Simply outputting the content would be:

val ls = io.Source.fromFile("Cg6367.scala").getLines 
println (ls.mkString ("\n"))

Answer 7

C, 135 chars

Though 40 of those 135 chars are just because there are some #includes that you just can't skip:

#include<sys/mman.h>
#include<sys/stat.h>

(Add fcntl.h and unistd.h for non-golfing completeness.) Then in the place where the file contents are desired, it's as simple as:

struct stat s;char*p;int f=open("filename.txt",0);fstat(f,&s);
p=mmap(0,s.st_size,1,2,f,0);close(f);

The file is closed and p contains a pointer to the file contents. (Don't forget to call munmap() when you're done.)

Answer 8

C

There's already a C answer, but this one uses nothing more than the absolute basics. It's also rather odd, and doesn't handle errors, because if users are going to give bad inputs really they deserve to have the program crash on them.

#include <stdlib.h>
#include <stdio.h>

**b,**m;main(l){
  FILE*f=fopen("foo.txt","r");
  for(m=b=malloc(64);~(l=fgetc(f));){
    *b++=&l;
    b=malloc(64);
  }
  return fclose(f);
}

Dumps the file into a linked list of chars, which if it's good enough for Haskell, it's good enough for C. Assumes a pointer is 32 bits and EOF is -1. Clearly a linked list of ints is the most

Answer 9

Ruby

• a=File.open('foo.txt', 'r') { |f| f.read }
• a=IO.read('foo.txt')

Answer 10

Clojure's slurp overload that doesn't take the encoding is deprecated (for good reason):

(slurp "a.txt")

Here's the way to specify the encoding

(slurp "a.txt" :encoding "UTF-8")