How to print out a backslash in LaTeX
Sound like you want a backslash in text mode; since \backslash does not work, try \textbackslash. EDIT: \symbol{92} should also work.
file-io
In golang how can I write the stdout of an exec.Cmd to a file?
Thanks to KirkMcDonald on the #go-nuts irc channel, I solved this by assigning the output file to cmd.Stdout, which means that stdout writes directly to the file. It looks like this: package main import ( “os” “os/exec” ) func main() { cmd := exec.Command(“echo”, “‘WHAT THE HECK IS UP'”) // open the out file for … Read more
How to read a file line by line in Julia?
Reading a file into memory all at once as an array of lines is just a call to the readlines function: julia> words = readlines(“/usr/share/dict/words”) 235886-element Array{String,1}: “A” “a” “aa” ⋮ “zythum” “Zyzomys” “Zyzzogeton” By default this discards the newlines but if you want to keep them, you can pass the keyword argument keep=true: julia> … Read more
What is the best variant for appending a new line in a text file?
Using OpenOptions::append is the clearest way to append to a file: use std::fs::OpenOptions; use std::io::prelude::*; fn main() { let mut file = OpenOptions::new() .write(true) .append(true) .open(“my-file”) .unwrap(); if let Err(e) = writeln!(file, “A new line!”) { eprintln!(“Couldn’t write to file: {}”, e); } } As of Rust 1.8.0 (commit) and RFC 1252, append(true) implies write(true). … Read more
How to read a binary file in Go
For manipulating files, the os package is your friend: f, err := os.Open(“myfile”) if err != nil { panic(err) } defer f.Close() For more control over how the file is open, see os.OpenFile() instead (doc). For reading files, there are many ways. The os.File type returned by os.Open (the f in the above example) implements … Read more
How can I “Save As…” in Emacs without visiting the new file?
Select the entire buffer with C-xh, then use M-xwrite-region.
How to create appending writeStream in Node.js
There is a createWriteStream method right below it. fs.createWriteStream(path, [options]) Returns a new WriteStream object (See Writable Stream). options is an object with the following defaults: { ‘flags’: ‘a’ , ‘encoding’: null , ‘mode’: 0666 }
How can I use the “compress/gzip” package to gzip a file?
All the compress packages implement the same interface. You would use something like this to compress: var b bytes.Buffer w := gzip.NewWriter(&b) w.Write([]byte(“hello, world\n”)) w.Close() And this to unpack: r, err := gzip.NewReader(&b) io.Copy(os.Stdout, r) r.Close()
Check if a file exists with Lua
Try function file_exists(name) local f=io.open(name,”r”) if f~=nil then io.close(f) return true else return false end end but note that this code only tests whether the file can be opened for reading.
Groovy write to file (newline)
As @Steven points out, a better way would be: public void writeToFile(def directory, def fileName, def extension, def infoList) { new File(“$directory/$fileName$extension”).withWriter { out -> infoList.each { out.println it } } } As this handles the line separator for you, and handles closing the writer as well (and doesn’t open and close the file each … Read more