How to print leading zeros (padding) in awk?

Question

I am trying to print 99.11111 as 099.11 in AWK.

I have tried the following variations without success.

$ awk '{printf ("%000.2f\n", $1);}' <<< 99.111111
99.11
$ awk '{printf ("%3.2f\n", $1);}' <<< 99.111111
99.11
$ awk '{printf ("%03.2f\n", $1);}' <<< 99.111111
99.11

Answer 1

Using any awk:

$ awk '{printf("%06.2f\n", $1)}' <<< 99.111111
099.11

The padding length takes into account the whole formatted number, so with the fractional part taking up 3 characters (. and the 2 precision digits), you need a padding length of 6 if you want the integral part to be padded to a length of 3. Note that the padding also takes into account the sign if any so -1 for instance would be formatted as -01.00¹

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^{¹ or possibly -01,00 or -01٫00 depending on the user's locale and the awk implementation². In the case of GNU awk, it also depends on whether there's a POSIXLY_CORRECT variable in the environment or not.}