Shell Programming and Scripting

I have a file as follows:

1
3
7
12
78
...
999998
1000000

I want to find out all numbers not in the file above and put into another file like

2
4
5
6
8
9
10
11
13
...
999999

thank you

One loop:

Code:

x=0 
c=0
 
while (( ++x <= 999999 ))
 do
  if (( x > c ))
   then
    read c
    if [ $? = 1 ]
     then
      c=1000000
     fi
   fi
  if (( x < c ))
   then
    echo $x
   fi
 done < infile

You need to check/generate all million numbers and then filter them with file numbers. Disposing of the edge cases tersely, never the same test twice, is important in getting a clean, minimal code. This ran almost 2 minutes for me. I could get smaller using the ternary operator '? :', but I am usually a ksh guy:

Code:

$ echo $(( 0 > 1 ? 2 : 3 ))
ksh:  0 > 1 ? 2 : 3 : syntax error
$ bash -c 'echo $(( 0 > 1 ? 2 : 3 ))'
3
$

Indentation indicates which line controls which. You, too, deserve properly indented code. It's an investment in your future.

If this is schoolwork, one needs to fill out a template.

What about

Code:

$ seq 1 1000000 | grep -vwf file

?

Isn't that an unordered search? Try:

Code:

$ seq 1 999999 | comm -23 - file

try also:

Code:

awk 'NR==1 {last=$1+1} {for (n=last; n<$1; n++) print n; last=$1+1}' infile

comm obviously expects an alphanumerically ordered file, so 10 after 9 makes it complain.
grep -w looks for entire words, so 10 doesn't match 100, and it doesn't care for order. seq (I hope) guarantees the order of the numbers.

Ah yes, need sorts:

Code:

comm -23 <( seq 1 999999 | sort ) <( sort file ) | sort -n

Less N x M. A numeric atof() option for comm might be nice.