Dynamic precision in awk printf using shell variable

Question

I think this is simple, but it's not working for me. This is what I have.

float=$(awk -v res="$result" 'BEGIN {printf "%.2f", res / 1000}')

I want to use a variable to set the decimal value for %.2f but awk will have none of it. This is what I need.

var=2
float=$(awk -v res="$result" 'BEGIN {printf "%.${var}f", res / 1000}')

I hope someone can show me the error of my ways.

Answer 1

awk's printf, like C's, lets you use a * instead of a hardcoded width or precision number. The value is instead taken from the next argument to the function. So instead of messing with shell substitution into the awk script, you can just pass the precision as another variable:

$ result=2050
$ prec=2
$ awk -v res="$result" -v p="$prec" 'BEGIN { printf "%.*f\n", p, res/1000 }'
2.05

---

However, if you're going to do a lot with floating point numbers, I'd suggest moving away from bash to another shell like zsh or ksh that supports floating-point math natively, or to a non-shell scripting language (perl, tcl, ruby, etc.). Tends to be more efficient than having to run external programs to do basic things like division.

Answer 2

I suggest:

result="123456"
var=2            # decimal places

awk -v res="$result" -v dp="$var" 'BEGIN {printf "%." dp "f", res / 1000}'

or

awk -v res="$result" -v CONVFMT="%.${var}f" 'BEGIN {printf res / 1000}'

Output:

123.46

From man awk:

CONVFMT: The conversion format for numbers, "%.6g", by default.