echo does not print a variable in bash

I’m facing a very strange issue. I have 2 files ber_log_before.txt and ber_log_after.txt. Their contents are

ber_log_before.txt

$
BOAR 0 BER:14
BOAR 1 BER:19
BOAR 2 BER:0
BOAR 3 BER:0
BOAR 4 BER:0
BOAR 5 BER:0
BOAR 6 BER:0
BOAR 7 BER:0

$

ber_log_after.txt

$
BOAR 0 BER:24
BOAR 1 BER:29
BOAR 2 BER:0
BOAR 3 BER:0
BOAR 4 BER:0
BOAR 5 BER:0
BOAR 6 BER:0
BOAR 7 BER:0

$

There are $ it’s normal.

Then I wrote a basic bash command to parse them

PORT="0 1"
for port in $PORT; do
    VAL1=$(grep "BOAR $port" ber_log_before.txt | cut -f2 -d':')
    VAL2=$(grep "BOAR $port" ber_log_after.txt  | cut -f2 -d':')
    echo 1st val ${VAL1} 2em val ${VAL2}
done

Why 1st val 14 2em val 24 is not echoed???

If I replace

echo 1st val ${VAL1} 2em val ${VAL2}

by

echo 1st val ${VAL1}
echo 2em val ${VAL2}

Then at least I have 2em val echoed.

I guess that awk can do the job, but that’s not the question.

Answer

This is not a problem actually, or it was me the problem. both files I mentioned are created from a telnet session (with expect command to be more specific). So all lines are ending with r and of course echo 1st val 14r 2em val 24r will ‘overwrite’ the first part of the echo. So I replace the r by n and then it’s ok.

To understand the problem, I copied my command into a script and use set -x. It was easy to find the r.

Sorry for that stupid question.