How can I make this bash code remove the decimal places?

Question

free | grep Mem | awk '{print $4/$2 * 100.0}'

I want to remove the decimal places from this output.

Answer 1

free | grep Mem | awk '{print $4/$2 * 100.0}' | cut -d. -f1

or

free | grep Mem | awk '{print int($4/$2 * 100.0)}'

Answer 2

A couple observations:

1. No need for using grep as awk can handle that
2. Use printf instead of print to control the precision

free | awk '/Mem/ { percent = $4 / $2 * 100.0; printf "%0.0f\n", percent; }'

Answer 3

1. Use awk /pattern/ to select the correct row.
2. Convert to integer to get rid of decimal points.

Code:

free | awk '/Mem/ { print int(100 * $4 / $2) }'