Bash – Why does arithmetic expansion work in Bash but not in Zsh

arithmeticbashshell-scriptzsh

I was demonstrating a piece of funny code (or so I think) to my students in class, and to my surprise, it did not work in zsh. I then switched to bash and it worked as expected.

The code is the following:

[ $(( $RANDOM % 6 )) == 0 ] && echo "Boom" || echo "Click"

In bash it prints Boom or Click randomly, as expected. But in zsh I get the error message below:

zsh: = not found

Can anyone explain what is the difference? To my understanding, both shells should interpret scripts in an equivalent manner.

Thanks.

Best Answer

Try doing unsetopt equals and try again. Or put quotes on the first equals character.

If you use [ for conditions, all arguments are evaluated normally and = at the start of a word is a useful zsh shorthand to get the location of a command. e.g. =ls expands to /bin/ls. Given == zsh looks for an = command and doesn't find one.

Zsh in posix mode (sh emulation) will be a lot more compatible with bash. Many things are compatible and most learnt knowledge of bash is still useful but there are certain things to be aware of if you want to write portable scripts.

Unlike [ which is a builtin, [[ is a reserved word understood by the parser and will give you fewer issues. However in this case you're using arithmetic mode so you simplify it:

    (( RANDOM % 6 == 0 )) && echo "Boom" || echo "Click"

In fact the comparison to zero is even superfluous because, the default return status reflects whether the value is zero, though the sense is reversed:

    (( RANDOM % 6 )) && echo "Click" || echo "Boom"

Related Solutions

Bash – Why does this command to copy files in a for loop work in bash but not in zsh

This is happening because cut is outputting NULL characters in the output. You can't pass a program arguments which contain a null character (see this).

In bash this works because bash can't handle NULL characters in strings, and it strips them out. Zsh is a bit more powerful, and it can handle NULL characters. However when it comes time to pass the string to the program, it still contains the null, which signals the end of the argument.

Let's look at this in detail.

$ echo 'English/Flavors/AU/stuff1/filename.txt' | cut -d'/' --output-delimiter '' -f1,3 | xxd
0000000: 456e 676c 6973 6800 4155 0a              English.AU.

Here we simulated one of your files, passing the path through cut. Notice the xxd output which has a NULL character between English and AU.

Now lets run through and simulate the rest of the script.

$ l=$(echo 'English/Flavors/AU/stuff1/filename.txt' | cut -d'/' --output-delimiter '' -f1,3)
$ dest=/stuff2/${l}.utf8.txt
$ echo "$dest" | xxd
0000000: 2f73 7475 6666 322f 456e 676c 6973 6800  /stuff2/English.
0000010: 4155 2e75 7466 382e 7478 740a            AU.utf8.txt.

Notice the NULL after the English. The echo displays it properly because echo is a shell built-in. If we use an external echo, it also exhibits the issue.

$ /bin/echo "$dest" | xxd 
0000000: 2f73 7475 6666 322f 456e 676c 6973 680a  /stuff2/English.

P.S. You really should be quoting too :-)

The solution is to not use cut, use awk instead.

$ echo 'English/Flavors/AU/stuff1/filename.txt' | awk -F/ '{ print $1$3 }' | xxd
0000000: 456e 676c 6973 6841 550a                 EnglishAU.

Bash – Why does this ffmpeg command work in bash and not zsh

In string:

rtsp://user:pass@my.webserver.org:5554/my-media/media.amp?videocodec=h264

you have ? in that string, so the shell will perform pathname expansion on that string, using pattern matching rules.

In bash, if failglob options was not set, which is default, then failed pattern will be left as-is:

$ echo does-not-exist?
does-not-exist?

While zsh will report no pattern match error with nomatch option set, which is default:

$ echo does-not-exist?
zsh: no matches found: does-not-exist?

You can make zsh suppress the error and print the pattern:

$ setopt nonomatch
$ echo does-not-exist?
does-not-exist?

You can make bash behave like zsh with nomatch option set, by turning on failglob:

$ shopt -s failglob
$ echo does-not-exist?
bash: no match: does-not-exist?

More general, you can disable shell filename generation:

$ set -f
$ : "The command"
$ set +f

(or set -o noglob, set +o noglob)

or using one of shell quoting methods to make the shell treats ? and other pattern matching special characters literally.

zsh also provide the noglob builtin, which disable filename generation in any words for the following simple command:

$ noglob echo *
*

Best Answer

Related Solutions

Bash – Why does this command to copy files in a for loop work in bash but not in zsh

Bash – Why does this ffmpeg command work in bash and not zsh

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