Copy list of files

No need for cat at all:

xargs -a list.txt cp -t new_folder

Or using long options:

xargs --arg-file=list.txt cp --target-directory=new_folder

Here are some shell versions. Note that some variables are unquoted or for loops are used specifically because the OP specified that the filenames are space delimited. Some of these techniques won't work for filename-per-line input in which filenames contain white space.

Bash:

for file in $(<list.txt); do cp "$file" new_folder; done

or

cp $(<list.txt) new_folder

sh (or Bash):

# still no cat!
while read -r line; do for file in $line; do cp "$file" new_folder; done; done < list.txt

or

# none here either
while read -r line; do cp $line new_folder; done < list.txt

or

# meow
for file in $(cat list.txt); do cp "$file" new_folder; done

or

# meow
cp $(cat list.txt) new_folder

Try using xargs:

cat list.txt | xargs -J % cp % new_folder

---

Update:

I did this on OS X which has an different version than GNU/Linux versions. The xargs which comes from GNU findutils doesn't have -J but it has -I which is similar (as Dennis Williamson pointed out in a comment). The OS X version of xargs has both -I and -J which have slightly different behaviors -- either will work for this original question.

$ cat list.txt
one
two.txt
three.rtf

$ cat list.txt | xargs -J % echo cp % new_folder
cp one two.txt three.rtf new_folder

$ cat list.txt | xargs -I % echo cp % new_folder
cp one new_folder
cp two.txt new_folder
cp three.rtf new_folder

I had an embarrassingly round-about answer here earlier, but Denis's answer reminded me that I missed the most basic thing. So, I deleted my original answer. But since nobody has said this very basic thing, I think it's worth putting it here.

The original question is "I have a text file with a list of space-separated names of files. How can I copy those to one target directory." At first this might seem tricky or complicated, because you think you have to somehow extract the items from the file in a specific way. However, when the shell processes a command line, the first thing it does is separate the argument list into tokens and (here's the bit nobody has said outright) spaces separate tokens. (Newlines also separate tokens which is why Doug Harris's test with a newline separated list had the same result.) That is, the shell expects and can already handle a space-separated list.

So all you need to do here is put the space-separated list (that you already have) into the right place in your command. Your command is some variation on this:

cp file1 file2 file3...file# target

The only wrinkle is that you want to get the list of files 1 through # from your text file.

As Dennis points out in his comment, your original attempt (cpcat list.txtnew_folder) should have worked already. Why? Because the internal command cat list.txt is processed first by the shell and expands into file1 file2 file3...file#, which is exactly what the shell expects and wants at that part of the command. If it didn't work then either (1) you had a typo or (2) your filenames were somehow strange (they had spaces or other unusual characters).

The reason that all Dennis's answers work is simply that they provide the necessary list of files for cp to work on, placing that list where it belongs in the entire command. Again, the command itself is this in structure:

cp list-of-files target_directory

It's easy to see how this all comes together in this version:

cp $(<list.txt) new_folder

$() causes the shell to run the command inside the parentheses and then substitute its output at that point in the larger line. Then the shell runs the line as a whole. By the way, $() is a more modern version of what you were already doing with backticks (`). Next: < is a file redirection operator. It tells the shell to dump the contents of list.txt to standard input. Since the $() bit gets processed first, here's what happens in stages:

1. cp $(<list.txt) new_folder # split line into three tokens: cp, $(<list.txt), new_folder
2. cp file1 file2 file3...file# new_folder # substitute result of $(<list.txt) into the larger command

Obviously step 2 is simply the normal cp command you wanted.

I realize that I'm beating this (perhaps very dead) horse a lot, but I think it's worth doing. Understanding exactly how the shell processes a command can help you to write it better and simplify a lot. It will also show you where problems are likely to be hiding. In this case, for example, my first question to you should have been about funny filenames or a possible typo. No acrobatics were necessary.