Bash variable expansion

Question

I'm trying to do some variable expansion in bash but somehow the result is truncated/rotated. Here's a sample my code:

x="no-cabac,level=3,ref=3,bframes=0,subme=0,weightp=0"
tts="{$x}"
echo $tts

This prints: }no-cabac,level=3,ref=3,bframes=0,subme=0,weightp=0

I expected: {no-cabac,level=3,ref=3,bframes=0,subme=0,weightp=0}

If I change tts to tts=abc{$x}qwe the output is }qweno-cabac,level=3,ref=3,bframes=0,subme=0,weightp=0

I tried escaping the braces and removing the quotes, still doesn't work. I know there's something that has to be escaped but I can't figure out what.

bash --version output: GNU bash, version 3.2.48(21)-release (i686-pc-cygwin)

That works fine for me with GNU bash, version 3.2.48(1)-release (x86_64-apple-darwin10.0) — R Samuel Klatchko
– R Samuel Klatchko, Commented Mar 7, 2010 at 19:42

Ignacio Vazquez-Abrams · Accepted Answer · 2010-03-07 19:49:10Z

5

You're getting your $x from somewhere else, and it has a \r at the end. Try:

tts="{${x/$'\r'/}}"

answered Mar 7, 2010 at 19:49

Ignacio Vazquez-Abrams

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3 Comments

Mauricio Scheffer Over a year ago

Thanks, that was it. I had my script file with CRLFs, therefore the extra \r :-/

Mauricio Scheffer Over a year ago

BTW I converted to LF and the original script worked just fine.

mklement0 Over a year ago

Just a diagnostic tip: pipe a variable value (or any string) to | od -a to see a character-by-character representation of it with whitespace and control characters showing with symbolic names. For instance, a \r shows as cr, \n as nl, \t as ht.

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