awk string parsing

I have an output that looks like this,

         1  2  3  4
5  6  7  8  9 10 11

Now I need to print every string ("1"..."11") plus some extra information. I tried a script like

BEGIN {FS = " "}
{
for (i = 1; i <= NF; i++)
{
   if ($i == myvar)
       printf("%s %s", $i, mystr)
   else
       printf("%s ", $i)
   if (i == NF)
       printf("\n")
}
}

This parses the first string("1") as appearing in column one, but it should appear in column 4, i.e the spaces in columns must be printed.

eing9607Asked:
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ahoffmannCommented:
do you mean the leading space left of 1 ?
And how about all the other leading spaces?
Then awk is probably not the rigth choice for that, but can be done too.
Could you please give an example (not phrases) how the output shoul look like.

Also, what does you program do ? the awk-variables myvar and mystr never get set, so they are always the empty string or 0 (depending on the context).

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eing9607Author Commented:
I haven't included the setting of myvar and mystr here, but I do set them in my awk script.

Well I kind of solved the problem myself. I use the split function, n = ($0, arr, " "), function to retrieve the number of elements on each line. Then I can print each entry stored in arr[1]..arr[n]. By doing so I can (pre)format the output the way I want.

Sorry, when I asked the question I wasn't aware of the split function.    
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