Re^2: typeglob: $var = *x ???

Thanks. So, I have to dereference the variable I assigned the typeglob to in order to turn it back into a typeglob? For instance,

$val = *x;
print "@{ *{ $val } }";
[download]

Based on the output here:

print $val;

--output:--
*main::x
[download]

...it's not obvious that $val needs dereferencing.

All three also work if $var = \*x;.

But I did $var = *x, which is not the same:

$what = *x;
print $what, "\n";

$what = \*x;
print $what, "\n";

*main::x
GLOB(0x10082ad88)
[download]

It seems like in the first case, I wouldn't have to dereference $what into a typeglob, where I would expect to do so in the second case.

Also, why can I do this:

$fh = *STDOUT;

print $fh 'hello', "\n";  

--output:--
hello
[download]

...and not this:

$STDOUT = 10;
$fh = *STDOUT;

print ($fh + 2);

--output:--
2
[download]

Comment on Re^2: typeglob: $var = *x ??? Select or Download Code

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Re^3: typeglob: $var = *x ??? by ikegami (Patriarch) on Jun 16, 2011 at 02:34 UTC
But I did $var = x* So did I. I said `\x` also* works. it's not obvious that $val needs dereferencing. I showed that it doesn't need dereferencing. `$gv = x; @$gv => @x` [download] Of course, one needs to use the sigil to access the components of the glob. `$gv = x; $gv{ARRAY} # Obviously wrong. $gv{ARRAY} # Ok` [download] ...and not this:* Do you really want `print($count + 2)` to use `$count` as a file handle? `print(STDOUT + 2) => print( { STDOUT } 2 )` [download] Barewords will be handled as expected, but anything else followed by what could be an infix operator will be treated as an argument to that operator. `print(STDOUT + 2) => print( { select() } STDOUT+2 ) print($x + 2) => print( { select() } $x+2 ) print($x + 2) => print( { select() } $x+2 )` [download]	[reply] [d/l] [select]
Re^4: typeglob: $var = *x ??? by 7stud (Deacon) on Jun 16, 2011 at 08:23 UTC
I showed that it doesn't need dereferencing. `$gv = *x; @$gv => @x` [download] perlreftut says that the syntax @$gv is a shortcut for @{$gv}, and that looks like the very definition of dereferencing to me (see perlreftut, Use Rule 1). As far as I can tell, the only time you don't have to dereference the variable is when you use it as a filehandle--in that case it appears that perl uses the context to determine that it should extract the filehandle part of the typeglob.	[reply] [d/l]
Re^5: typeglob: $var = *x ??? by ikegami (Patriarch) on Jun 16, 2011 at 15:30 UTC
that looks like the very definition of dereferencing to me I thought you meant `$gv` vs `$gv` it's not obvious that $val needs dereferencing.* Completely false. "@foo" returns the list of elements in the array. "$foo" does not. According to you, (A) and (B) and (C) should be the same? `A: @m = $x; B: @m = @$x; C: @m = $$x;` [download] The only thing that may not be obvious is that globs can be "dereferenced" instead of having to do `$gv{ARRAY}`. As far as I can tell, the only time you don't have to dereference the variable is when you use it as a filehandle* Both the "`@`" and "`print`" operators accept a scalar containing a glob. The first peeks into the ARRAY slot. The latter peeks into the IO slot. Both "dereference" the glob.	[reply] [d/l] [select]