Avoiding lc on a reference value

[Tony]

Is there an elegant way to determine if a variable contains a plain
old value?

I'm working through an array, and lowercasing everything. I need to
skip over elements that are references, and the below seems a bit
ugly/prone to error:


if ($components[$i] !~ /(0x|SCALAR)/) {
$components[$i] = lc($components[$i]);
}


Thank you!

 

[Dmitry Roslyakov]

Is there an elegant way to determine if a variable contains a plain
old value?

I'm working through an array, and lowercasing everything. I need to
skip over elements that are references, and the below seems a bit
ugly/prone to error:


if ($components[$i] !~ /(0x|SCALAR)/) {
$components[$i] = lc($components[$i]);
}


Thank you!

$components[$i] = lc($components[$i]) unless ref($components[$i]);

 

[David K. Wall]

Is there an elegant way to determine if a variable contains a plain
old value?

I'm working through an array, and lowercasing everything. I need to
skip over elements that are references, and the below seems a bit
ugly/prone to error:


if ($components[$i] !~ /(0x|SCALAR)/) {
$components[$i] = lc($components[$i]);
}

I think you want ref(). See

perldoc -f ref

 

[Uri Guttman]

Is there an elegant way to determine if a variable contains a plain
old value?
I'm working through an array, and lowercasing everything. I need to
skip over elements that are references, and the below seems a bit
ugly/prone to error:
if ($components[$i] !~ /(0x|SCALAR)/) {
$components[$i] = lc($components[$i]);
}
Thank you!

DR> $components[$i] = lc($components[$i]) unless ref($components[$i]);

GACK!!!

i will have to assume a c style loop in the OP's code there which is not
perlish

ref or $_ = lc for @components ;

uri