How to read first and last line from cat output?

sed Solution:

sed -e 1b -e '$!d' file

When reading from stdin if would look like this (for example ps -ef):

ps -ef | sed -e 1b -e '$!d'
UID        PID  PPID  C STIME TTY          TIME CMD
root      1931  1837  0 20:05 pts/0    00:00:00 sed -e 1b -e $!d

head & tail Solution:

(head -n1 && tail -n1) <file

When data is coming from a command (ps -ef):

ps -ef 2>&1 | (head -n1 && tail -n1)
UID        PID  PPID  C STIME TTY          TIME CMD
root      2068  1837  0 20:13 pts/0    00:00:00 -bash

awk Solution:

awk 'NR==1; END{print}' file

And also the piped example with ps -ef:

ps -ef | awk 'NR==1; END{print}'
UID        PID  PPID  C STIME TTY          TIME CMD
root      1935  1837  0 20:07 pts/0    00:00:00 awk NR==1; END{print}

sed -n '1p;$p' file.txt will print 1st and last line of file.txt .

A funny pure Bash≥4 way:

cb() { (($1-1>0)) && unset "ary[$1-1]"; }
mapfile -t -C cb -c 1 ary < file

After this, you'll have an array ary with first field (i.e., with index 0) being the first line of file, and its last field being the last line of file. The callback cb (optional if you want to slurp all lines in the array) unsets all the intermediate lines so as to not clutter memory. As a free by-product, you'll also have the number of lines in the file (as the last index of the array+1).

Demo:

$ mapfile -t -C cb -c 1 ary < <(printf '%s\n' {a..z})
$ declare -p ary
declare -a ary='([0]="a" [25]="z")'
$ # With only one line
$ mapfile -t -C cb -c 1 ary < <(printf '%s\n' "only one line")
$ declare -p ary
declare -a ary='([0]="only one line")'
$ # With an empty file
$ mapfile -t -C cb -c 1 ary < <(:)
declare -a ary='()'