CODEWITHSUNDEEP
bash
MathematicalOrchid
MathematicalOrchid

Reputation: 62818

Failure of Bash $@ variable

I've done this several times, and it never seems to work properly. Can anyone explain why?

function Foobar
{
  cmd -opt1 -opt2 $@
}

What this is supposed to do is make it so that calling Foobar does the same thing as calling cmd, but with a few extra parameters (-opt1 and -opt2, in this example).

Unfortunately, this doesn't work properly. It works OK if all your arguments lack spaces. But if you want an argument with spaces, you write it in quotes, and Bash helpfully strips away the quotes, breaking the command. How do I prevent this incorrect behavior?

Upvotes: 2

Views: 110

Answers (1)

Gordon Davisson
Gordon Davisson

Reputation: 125798

You need to double-quote the $@to keep bash from performing the unwanted parsing steps (word splitting etc) after substituting the argument values:

function Foobar
{
  cmd -opt1 -opt2 "$@"
}

EDIT from the Special Parameters section of the bash manpage:

@   Expands to the positional parameters, starting from  one.   When
    the  expansion  occurs  within  double  quotes,  each  parameter
    expands to a separate word.  That is, "$@" is equivalent to "$1"
    "$2"  ...   If the double-quoted expansion occurs within a word,
    the expansion of the first parameter is joined with  the  begin-
    ning  part  of  the original word, and the expansion of the last
    parameter is joined with the last part  of  the  original  word.
    When  there  are no positional parameters, "$@" and $@ expand to
    nothing (i.e., they are removed).

Upvotes: 8

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