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Linux: execution time to 1000th of a second

#!/bin/bash
t1=$( date +"%s" )
sleep 1.23
t2=$( date +"%s" )
DIFF=$(( $t2 - $t1 ))
echo "Exicution time: $DIFF seconds"

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I want the results to be like
1.234 seconds
0
hankknight
Asked:
hankknight
  • 4
1 Solution
 
jb1devCommented:
You need to add %N to your date format for nanoseconds and you need to use bc for floating point evaluation rather than just expr "-"

#!/bin/bash
t1=$( date +"%s.%N" )
sleep 1.23
t2=$( date +"%s.%N" )
echo "$t2 - $t1"
DIFF=$(echo "($t2 - $t1) * 1000" | bc)
echo "Execution time: $DIFF seconds"

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See also
<Link removed by woolmilkporc, EE TA>
0
 
jb1devCommented:
Sorry that *1000 is incorrect.

That should be

#!/bin/bash
t1=$( date "+%s.%N" )
sleep 1.23
t2=$( date "+%s.%N" )
echo "$t2 - $t1"
DIFF=$(echo "( $t2 - $t1 )" | bc)
echo "Exicution time: $DIFF seconds"

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You can then add scale to bc to limit this to 3 digits. (or however many.)
0
 
jb1devCommented:
Here you go with scale set to your desired precision

#!/bin/bash
t1=$( date "+%s.%N" )
sleep 1.23
t2=$( date "+%s.%N" )
echo "$t2 - $t1"
DIFF=$(echo "scale=3; ($t2 - $t1)/1" | bc)
echo "Execution time: $DIFF seconds"

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EDIT Note scale does not apply to subtraction. See this <Link removed by woolmilkporc, EE TA> so we divide by 1 (with /1) in order for the floating point scale to be applied.
0
 
jb1devCommented:
@woolmilkporc
Sorry about that. I have not been posting here in a while.
I was not aware of this rule.
0
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