In bash, how can I print the first n elements of a list?

Question

In bash, how can I print the first n elements of a list?

For example, the first 10 files in this list:

FILES=$(ls)

Answer 1

FILES=(*)
echo "${FILES[@]:0:10}"

Should work correctly even if there are spaces in filenames.

FILES=$(ls) creates a string variable. FILES=(*) creates an array. See this page for more examples on using arrays in bash. (thanks lhunath)

Answer 2

echo $FILES | awk '{for (i = 1; i <= 10; i++) {print $i}}'

Edit: AAh, missed your comment that you needed them on one line...

echo $FILES | awk '{for (i = 1; i <= 10; i++) {printf "%s ", $i}}'

That one does that.