Bash – How to cat a file I got from stdin without xargs

I recommend reading a book on unix or Linux shell and command line usage, in order to learn basic usage and get a feeling for some advanced features. Then you can turn to reference documentation.

The usage of specific commands is described in their manual. man cat will show the manual of the cat command on your system. Manual pages are usually references, not tutorials, though they often contain examples. On Linux, cat --help shows a terse usage message (meant for quick perusal when you already know the fundamentals and want to find an option for a specific task).

The POSIX standard specifies a minimum set of commands, options and shell features that every unix system is supposed to support. Most current systems by and large support POSIX:2004 (also known as Single UNIX version 3 and the Open Group Base Specifications issue 6). GNU software (the utilities found on Linux) often have many extensions to this minimum set.

There are common conventions for command-line arguments. POSIX specifies utility conventions that most utilities follow, in particular:

• Options consist of - followed by a single letter; -ab is shorthand for -a -b.
• -- signifies the end of options. For example, in rm -- -a, -a is not an option but an operand, i.e. a file to act upon, so this commands removes the file called -a.
• A lone - stands for standard input, where an input file is expected. It stands for standard output where an output file is expected.

GNU utilities and others also support “long options” of the form --name. Some utilities go against the general convention and take multi-letter options with a single leading dash: -name.

Redirection is a shell feature, so you'll find it in your shell's manual. <<< to use a string as standard input is a ksh extension, also supported by bash and zsh. As long as the shell supports it, it can be used on any command.

First of all, do not use ls output as a file list. Use shell expansion or find. See below for potential consequences of ls+xargs misuse and an example of proper xargs usage.

1. Simple way: for loop

If you want to process just the files under A/, then a simple for loop should be enough:

for file in A/*.dat; do ./a.out < "$file" > "${file%.dat}.ans"; done

2._pre1 Why not   ls | xargs ?

Here's an example of how bad things may turn if you use ls with xargs for the job. Consider a following scenario:

• first, let's create some empty files:

  $ touch A/mypreciousfile.dat\ with\ junk\ at\ the\ end.dat
  $ touch A/mypreciousfile.dat
  $ touch A/mypreciousfile.dat.ans
  

• see the files and that they contain nothing:

  $ ls -1 A/
  mypreciousfile.dat
  mypreciousfile.dat with junk at the end.dat
  mypreciousfile.dat.ans
  
  $ cat A/*
  

• run a magic command using xargs:

  $ ls A/*.dat | xargs -I file sh -c "echo TRICKED > file.ans"
  

• the result:

  $ cat A/mypreciousfile.dat
  TRICKED with junk at the end.dat.ans
  
  $ cat A/mypreciousfile.dat.ans
  TRICKED
  

So you've just managed to overwrite both mypreciousfile.dat and mypreciousfile.dat.ans. If there were any content in those files, it'd have been erased.

2. Using  xargs : the proper way with  find 

If you'd like to insist on using xargs, use -0 (null-terminated names) :

find A/ -name "*.dat" -type f -print0 | xargs -0 -I file sh -c './a.out < "file" > "file.ans"'

Notice two things:

1. this way you'll create files with .dat.ans ending;
2. this will break if some file name contains a quote sign (").

Both issues can be solved by different way of shell invocation:

find A/ -name "*.dat" -type f -print0 | xargs -0 -L 1 bash -c './a.out < "$0" > "${0%dat}ans"'

3. All done within find ... -exec

 find A/ -name "*.dat" -type f -exec sh -c './a.out < "{}" > "{}.ans"' \;

This, again, produces .dat.ans files and will break if file names contain ". To go about that, use bash and change the way it is invoked:

 find A/ -name "*.dat" -type f -exec bash -c './a.out < "$0" > "${0%dat}ans"' {} \;