I have this code :

tr ' ' '\n'|sort -n|head -1

The code is pretty simple, it takes a list of integer in the standard input stream (like 5 9 8 7 5 2 -12 -30), it sort the integers and display only the first integer. But i need to ouput 0 if the given list is empty. And i can't find a short way to that. I have a solution with a ternary operator and a echo 0, but i feel i can do far better. I just need to replace the input stream by 0 if this stream is empty.

Sidenote : I don't need to call read to get the user input. The program is called with the input stream directly like cmd < 1 5 6 -20 45

Here my current code :

read l
[ -n "$l" ]&&echo $l|tr ' ' '\n'|sort -n|head -1||echo 0
  • \$\begingroup\$ @Pietu1998 i added my current code \$\endgroup\$
    – Magus
    Feb 15, 2016 at 13:16
  • \$\begingroup\$ In this case you can cheat by always outputting a 0 first, if leading zeroes are allowed. \$\endgroup\$
    – orlp
    Feb 15, 2016 at 13:47
  • \$\begingroup\$ @orlp nice thinking but sadly leading zeroes are not allowed \$\endgroup\$
    – Magus
    Feb 15, 2016 at 13:59
  • \$\begingroup\$ This looks like a programming problem to go on Stack Overflow. I'm not sure though; I haven't done any action on the question in terms of close/up/down voting. \$\endgroup\$
    – hyper-neutrino
    Feb 15, 2016 at 19:40
  • \$\begingroup\$ @AlexL. Make code X shorter will be very poorly received on Stack Overflow. \$\endgroup\$
    – Dennis
    Feb 15, 2016 at 19:41

4 Answers 4


34 bytes

a=($(tr \  \\n|sort -n))
echo $[a]

The first line saves the sorted input values in an array a.

This avoids using head -1, since referencing the array as $a will yield its first value.

The second line uses a in the arithmetic expansion $[a]. In this context, an empty string is interpreted as 0.

For a different default value, parameter substitution could be used instead. For example, the last line could become

echo ${a:-puppies}

37 33 32 bytes

tr \  \\n|sort -n|sed s/^$/0/\;q

Edit: Saved 4 bytes thanks to @DigitalTrauma.

  • \$\begingroup\$ sed 1q is one shorter than head -1. Then you can combine the sed expressions to get 33 bytes: tr \ \\n|sort -n|sed s/^$/0/\;1q \$\endgroup\$ Feb 15, 2016 at 17:57
  • 1
    \$\begingroup\$ @DigitalTrauma Am I overlooking something or does sed q work? \$\endgroup\$
    – Neil
    Feb 15, 2016 at 19:24
  • \$\begingroup\$ yes - even better! \$\endgroup\$ Feb 15, 2016 at 19:25
  • \$\begingroup\$ @DigitalTrauma @Neil The sed script can be shortened by 1 byte further using /./q\;c\0 as long as the empty list doesn't contain white spaces. \$\endgroup\$
    – seshoumara
    Aug 29, 2016 at 14:48
  • 1
    \$\begingroup\$ tr \ \\n -> fmt -1 \$\endgroup\$
    – user98766
    Oct 17, 2020 at 4:11

42 bytes

Two different ways that have the same byte-count Use awk as the last command

tr ' ' '\n'|sort -n|awk 'NR==1{print $1+0}'

Or do all the computations in awk

awk -v RS=" " '{m=m>$1?m:$1}END{print m+0}'
  • 1
    \$\begingroup\$ I could shrink the all in awk version by a couple bytes using awk -v RS=" " '$1>m{m=$1}END{print m+0}', but then the two versions would have different byte-counts. :p \$\endgroup\$ Feb 15, 2016 at 15:08

Pure : 41 bytes.

No requirement of any tr, sort, head...

t=${1:-0};for i;{ ((t=i<t?i:t));};echo $t

In a function you could use local -i for integer variables:

f() { local -i t=$1;for i;{ ((t=i<t?i:t));};echo $t;}


f 5 9 8 7 5 2 -12 -30


Note: If you need to read them from stdin

read l
f $l

will do the job.


read l;for i in ${l:-0};{ ((t=i<t?i:t));};echo $t;}

Another version 58 bytes.

Something longer, but could be quicker, as there is no loop.

. <(printf v[%d+99**9]=\  $@);v=(${!v[@]});echo $[v-99**9]

In a function:

f() { local v;. <(printf v[%d+99**9]=\  $@)
      v=(${!v[@]});echo $[v-99**9];}


f 34 56 1000 12  -156 53 -12 -21 321 3221


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