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From: Bart Schaefer <schaefer@brasslantern.com>
Message-id: <120221214137.ZM13624@torch.brasslantern.com>
Date: Tue, 21 Feb 2012 21:41:37 -0800
In-reply-to:
 <CAPg-njyehwyETKY4KGdFV8u_ZwkMku-G9xVi7d7PpU9rZhdPDA@mail.gmail.com>
Comments: In reply to Daniel Lin <dlin.tw@gmail.com>
 "let unset array element remove compatible with bash" (Feb 22,  3:28am)
References: <CAPg-njyehwyETKY4KGdFV8u_ZwkMku-G9xVi7d7PpU9rZhdPDA@mail.gmail.com>
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To: zsh-users@zsh.org
Subject: Re: let unset array element remove compatible with bash
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On Feb 22,  3:28am, Daniel Lin wrote:
}
} Can any developer consider to enhance zsh's function like "unset var[2]"?
} 
}   $unset var[2] ###### BASH only delete one element
}   $var[2]=() ###### ZSH only delete one element

I've started a thread on zsh-workers about this, but:

var[3]=() does not mean the same thing that unset var[3] would imply.

In zsh, if you assign to a position that is "off the end" of the array,
zsh manufactures empty array elements to "fill in the gap".  Try this
in each of bash and zsh:

    unset gappy
    gappy[9]=nine
    for g in "${gappy[@]}"; do echo /$g/; done

Now in zsh:

    unset gappy
    gappy=(one)
    gappy[3]=()
    for g in "${gappy[@]}"; do echo /$g/; done

Note that assigning an empty array to the one-element slice gappy[3]
has caused gappy[2] to exist as an empty element.

Bash is using some kind of sparse structure to store its arrays.