elapsed function in bash to print how long things take

I needed this for a project and it has served me pretty well. Let's jump right into it:

# This is elapsed.sh

SECONDS=0

function elapsed()
{
  local T=$SECONDS
  local D=$((T/60/60/24))
  local H=$((T/60/60%24))
  local M=$((T/60%60))
  local S=$((T%60))
  (( $D > 0 )) && printf '%d days ' $D
  (( $H > 0 )) && printf '%d hours ' $H
  (( $M > 0 )) && printf '%d minutes ' $M
  (( $D > 0 || $H > 0 || $M > 0 )) && printf 'and '
  printf '%d seconds\n' $S
}

And here's how you use it:

# Assume elapsed.sh to be in the current working directory
source elapsed.sh

echo "Doing some stuff..."
# Imagine it does something slow that
# takes about 3 seconds to complete.
sleep 3
elapsed

echo "Some quick stuff..."
sleep 1
elapsed

echo "Doing some slow stuff..."
sleep 61
elapsed

The output of running that is:

Doing some stuff...
3 seconds
Some quick stuff...
4 seconds
Doing some slow stuff...
1 minutes and 5 seconds

Basically, if you have a bash script that does a bunch of slow things, it having a like of elapsed there after some blocks of code will print out how long the script has been running.

It's not beautiful but it works.