Append files with bash one-liner?

(./myprogram 1.txt; ./myprogram 2.txt; ./myprogram 3.txt; ./myprogram 4.txt) > out.txt

It depends on what you want to do and how your program handle input args.

But let's say you have /path/to/myprogram.sh who look like this one :

#!/bin/bash
echo "Processing file ${1?missing input file}"

You could do the following

find /path/to/inputfiles -name "*.txt" -exec /path/to/myprogram.sh {} \; > Out.txt

Or within bash (or any Bourne-like shell):

for i in *.txt; do /path/to/myprogram.sh "$i"; done > Out.txt

(I use for-loop or find because it will be much more convenient if you have 1000 input files instead of the 4 files in your example)

Very similar to Per's answer, but maintains the layout of the original:

{
    ./myprogram 1.txt
    ./myprogram 2.txt
    ./myprogram 3.txt
    ./myprogram 4.txt
} > Out.txt

http://www.gnu.org/software/bash/manual/bashref.html#Command-Grouping

myprogram-wrapper.sh:

#!/bin/sh -
while test $# -ge 1
do
  ./myprogram $1
  shift
done

invoke with ./myprogram-wrapper.sh 1.txt 2.txt 3.txt 4.txt > Out.txt

To make a reusable wrapper:

#!/bin/sh -
prog=$1
shift
while test $# -ge 1
do
  ${prog} $1
  shift
done

and invoke like wrapper ./mycommand 1.txt 2.txt 3.txt 4.txt > Out.txt

This is of course cheating, since the script itself is not a oneliner, but I thought you might like it anyway.

I find this solution more elegant:

FILE=/tmp/file.log

exec >> $FILE
./myprogram 1.txt
./myprogram 2.txt
./myprogram 3.txt
./myprogram 4.txt
exec 1<&2

echo 'Normal output has returned!'