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Editorial Team
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Asked: 2026-06-05T11:28:29+00:00

I’m writing a quick script for calculating individual interface throughput from data pulled from

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I’m writing a quick script for calculating individual interface throughput from data pulled from /proc/net/dev and I’m having an issue. It converts it from bytes to megabits.

This is working on my ubuntu server (3.2.0 kernel), however when I try and run this on older devices (2.6.18 era) it’s not working. I’m not sure what I’m doing wrong.

Here’s a snippet of my code:

int1_byte_rx=`cat $logfile | grep $int1 | awk '{print $2}' | awk '{sum+=$1} END {print sum}'`
int1_byte_tx=`cat $logfile | grep $int1 | awk '{print $10}' | awk '{sum+=$1} END {print sum}'`
int1_rx_thrpt=$(echo "($int1_byte_rx * 0.00000762939453) / $iterations / ($time * 60)" | bc -l)
int1_tx_thrpt=$(echo "($int1_byte_tx * 0.00000762939453) / $iterations / ($time * 60)" | bc -l)

When I run this I get the following error (from debug mode):

int1_rx_thrpt=$(echo "($int1_byte_rx * 0.00000762939453) / $iterations / ($time * 60)" | bc -l)
echo "($int1_byte_rx * 0.00000762939453) / $iterations / ($time * 60)" | bc -l
++ echo '(1.13417e+10 * 0.00000762939453) / 57 / (5 * 60)'
++ bc -l
(standard_in) 1: parse error
(standard_in) 1: parse error
+ int1_rx_thrpt=
int1_tx_thrpt=$(echo "($int1_byte_tx * 0.00000762939453) / $iterations / ($time * 60)" | bc -l)
echo "($int1_byte_tx * 0.00000762939453) / $iterations / ($time * 60)" | bc -l
++ echo '(9.78048e+09 * 0.00000762939453) / 57 / (5 * 60)'
++ bc -l
(standard_in) 1: parse error
(standard_in) 1: parse error

I’ve been able to trace the problem back to bc itself, however I’m not actually sure how to correct it.

Any advice is welcome.

Thanks for your time,

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1 Answer

  1. Editorial Team

    My /proc/net/dev doesn’t have large enough values to force awk to print any output in scientific notation, so I cannot easily test this, but here’s my suggested fix:

    int1_byte_rx=`cat $logfile | grep $int1 | awk '{print $2}' | awk '{sum+=$1} END {printf "%f", sum}'`
int1_byte_tx=`cat $logfile | grep $int1 | awk '{print $10}' | awk '{sum+=$1} END {printf "%f", sum}'`

    (Note the printf "%f", portion near the end.)

    The trick is to prevent awk from generating the scientific format in the first place.

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