Re: [Help-bash] declare -i

[Dan Douglas]

On Wednesday, May 09, 2012 11:24:36 AM Bill Gradwohl wrote:
> 
> > On Tue, May 8, 2012 at 10:03 PM, Dan Douglas <address@hidden> wrote:
> > 
> > Using -i can be an easy source of bugs and security problems, and adds a 
lot
> > of
> > quirks especially with arrays.
> > 
> > -- Dan Douglas
> 
> 
> Is this one of the quirks you were talking about:
> 
> declare -i -a intArray=(' 5' ' 2 ' 1 '  -19  ' 'xyz')

Yes that's one of them. -i gets set after the argument(s) are processed. This 
is yet another thing affected by everything disucssed in your "indirection" 
thread.

 $ declare -ai a=(1 1+1 3); declare -p a
declare -ai a='([0]="1" [1]="1+1" [2]="3")'

 $ declare -ai 'a=(1 1+1 3)'; declare -p a
declare -ai a='([0]="1" [1]="2" [2]="3")'

And it's also required to modify a global array:

 $  a=(1 2 3); f() { local a=(a b c); declare -ga 'a+=(4)'; declare -p a; }; 
f; declare -p a
 declare -a a='([0]="a" [1]="b" [2]="c")'
 declare -a a='([0]="1" [1]="2" [2]="3" [3]="4")'

But this is probably a bug:

 $ a=(1 2 3); f() { local a=(a b c); declare -ga a+=(4); declare -p a; }; f; 
declare -p a
 -bash: declare: `a+': not a valid identifier
 declare -a a='([0]="a" [1]="b" [2]="c" [3]="4")'
 declare -a a='([0]="1" [1]="2" [2]="3")'

Most of the anomolies aren't necessarily "bugs", just undocumented or conflict 
with documented things. One is that += can unset the zeroth element in Bash if 
the index isn't given explicitly. += doesn't unset elements in any other case.

 $ declare -ia a=(2 2 3); a+=1; declare -p a
 declare -ai a='([0]="1" [1]="2" [2]="3")'

Ksh doesn't have this problem:

 $ typeset -ia a=(2 2 3); a+=1; typeset -p a
 typeset -a -i a=(3 2 3)

Another the manual doesn't mention is that negative indexes can't be used on 
assignment, even if the array already exists (ksh can, but zsh and mksh 
completely disagree here too):

 $ declare -ia a=(1 2 3); read -r 'a[-1]' <<<'a[-1]+1'; declare -p a
-bash: a[-1]: bad array subscript
declare -ai a='([0]="1" [1]="2" [2]="3")'

 $ typeset -ia a=(1 2 3); read -r 'a[-1]' <<<'a[-1]+1'; typeset -p a
typeset -a -i a=(1 2 4)

Bash is the only shell that evaluates expansions in an array index indirectly, 
so getting strings into your integer array can be dangerous. Here's a forkbomb 
that would probably stump most people (careful):

declare -i _=('_[`_=_&`]')

-- 
Dan Douglas

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