File input/output: Difference between revisions

This task is easy, with 'fileutils' from the standard library.

<lang ruby>require 'fileutils'

FileUtils.copy_file 'input.txt', 'output.txt'</lang>

It also works with open IO handles.

<lang ruby>File.open('input.txt', 'rb') do |i|

File.open('output.txt', 'wb') do |o|

FileUtils.copy_stream(i, o)

end

end</lang>

----

We can also copy a file with only the core language.

<lang ruby>File.open('input.txt', 'rb') do |i|

File.open('output.txt', 'wb') do |o|

buf = ""

bufsiz = (i.stat.blksize or 16384)

while i.read(bufsiz, buf) do o.write(buf) end

end

end</lang>

* The best buffer size is <code>i.stat.blksize</code>. Some platforms return <code>nil</code> there, so we guess 16384 bytes.

The shortest way to copy a file, without coding any loops and without requiring any libraries, is to read the entire file into memory.

<lang ruby>irb(main):001:0> open('output.txt', 'w') {|f| f << IO.read('input.txt')}

=> #<File:output.txt (closed)></lang>

But this has disadvantages:

* There was no 'b' flag, so it might not work with binary files on some platforms. With a 'b' flag, the code would be <code>open('output.txt', 'wb') {|f| f << IO.read('input.txt', mode: 'rb')}</code>

* If the file is very large, we fail to allocate enough memory.

----

Ruby 1.9 has a new core method, <code>IO.copy_stream</code>. With [[MRI]] 1.9, <code>IO.copy_stream</code> might use [[Linux]]'s [http://www.kernel.org/doc/man-pages/online/pages/man2/sendfile.2.html sendfile(2)] system call, or it might loop with a buffer of 16384 bytes.

{{works with|Ruby|1.9}}

<lang ruby>IO.copy_stream('input.txt', 'output.txt')

# It also works with open IO handles.

File.open('input.txt', 'rb') do |i|

File.open('output.txt', 'wb') do |o|

IO.copy_stream(i, o)

end

end</lang>

FileUtils knows about IO.copy_stream; so if you use FileUtils, then your program uses IO.copy_stream with Ruby 1.9, but still works with older Ruby versions.