Re^2: hash parameter question

That's exactly what's wrong
my %user=$_;
should be
my %user=%$_;
because $_ contains a *reference* to the nested hash. The fact that a scalar is being assigned to a hash is a sure sign something is wonky.

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Re^3: hash parameter question by jarich (Curate) on Dec 13, 2005 at 02:43 UTC
Nope. `$_` contains a string. A key from `%users`; which is presumably a username. Your suggestion will therefore not work. Had PerlHeathen written: `foreach(values %users) { my %user=$_; my $pw=getpwnam($user{uname}); my $homedir=$pw->dir; }` [download] (note the use of values) then $_ would be a reference to a hash and then your solution would make the code work. For example: `foreach(values %users) { my %user=%$_; my $pw=getpwnam($user{uname}); my $homedir=$pw->dir; }` [download] It is likely that this is what PerlHeathen was trying to write in the first place. jarich	[reply] [d/l] [select]
Re^4: hash parameter question by PerlHeathen (Initiate) on Dec 14, 2005 at 19:12 UTC
Thank you all. This is what I was looking for; I need both suggestions -- pass by reference & use values(%users) instead. thanks again!	[reply]