Re^4: using shift and $

Don't clobber your caller's variables.. so.. in ..

sub a {
 @_;
 $_;
}
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$_ is .. what of the caller?

And am I to understand that in sub a, messing with @_, regardless if the caller passed any references as arguments, is potentially messing with the caller's arguments?

(thank you for your patience here)

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Re^5: using shift and $_ in loop by ikegami (Patriarch) on Oct 21, 2009 at 20:34 UTC
$_ is .. what of the caller? `$_` is `$_`. No telling how the caller is using it. And am I to understand that in sub a, messing with @_, regardless if the caller passed any references as arguments, is potentially messing with the caller's arguments? Easy to test... Not quite. Messing with the elements of `@_` will. Adding elements and removing elements (including assigning to `@_` specifically) won't. `foo($x, $y+1, $z);` [download] is basically equivalent to `{ local @_; alias $_[0] = $x; alias $_[1] = $y+1; alias $_[2] = $z; &foo; }` [download] Changing `$_[0]` in `foo` will change `$x`, for example.	[reply] [d/l] [select]

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Re^5: using shift and $_ in loop
by ikegami (Patriarch) on Oct 21, 2009 at 20:34 UTC

$_ is .. what of the caller?

$_ is $_. No telling how the caller is using it.

And am I to understand that in sub a, messing with @_, regardless if the caller passed any references as arguments, is potentially messing with the caller's arguments?

Easy to test...

Not quite. Messing with the elements of @_ will. Adding elements and removing elements (including assigning to @_ specifically) won't.

foo($x, $y+1, $z);
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{
    local @_;
    alias $_[0] = $x;
    alias $_[1] = $y+1;
    alias $_[2] = $z;
    &foo;
}
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Changing $_[0] in foo will change $x, for example.

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