Sort entries of lines using shell

Question

Considering the following input and output:

  infile   |   outfile
1 3 5 2 4  |  1 2 3 4 5
2 4 5      |  2 4 5
4 6 2 1    |  1 2 4 6

Is there a

Answer 1

Its not pretty (definitely not a 1-liner), but you can sort a line using only builtin shell commands, however for short lines it may be faster than repeatedly calling external functions.

#!/bin/sh
sortline(){
for x in $@;do
    [ ! "$FIRST" ] && FIRST=t && set --
    i=0
    while [ $i -le $# ];do
        [ $x -lt $((${@:$((i+1)):1})) ] && break || i=$((i+1))
    done
    set -- ${@:1:$i}  $x   ${@:$((i+1)):$(($#-$i))}
done
echo $@
}
while read LINE || [ "$LINE" ];do
    sortline $LINE
done <$1 >$2

Edit: btw this is a selection sort algorithm in case anyone wondered

Edit2: this is for numerical values only, for strings you would need to use some comparison like [ "$x" -lt "${@:$((i+1)):1}" ] (unchecked),however I use this C program for strings (I just call it qsort), but it could be modified using atoi on argv:

#include 
#include 
static inline int cmp(const void *a, const void *b){
   return strcmp(*(const char **)a, *(const char **)b);
}

int main(int argc, char *argv[]){
    qsort(++argv, --argc, sizeof(char *), cmp);
    while (argc){
      write(1,argv[0],strlen(argv[0]));
      write(1,(--argc && argv++)?"\t":"\n",1);
   }
}