Just as an academic exercise, I thought I should be able to do this:

**************************************
[at] a=(l=>35,k=>31,r=>7,k=>6);
[at] r=qw/l r r k/;
# make an anonymous hash using [at] a, then grab values from it using [at] r as keys
[at] a={ [at] a}{ [at] r};
print join(",", [at] a), "\n"'
**************************************

.... but it doesn't like the {}{} notation. Is this not possible? I do this
with lists all the time, e.g. [at] a = (split)[1,7,15];

TIA.

- Bryan



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>>>>> "BRH" == Bryan R Harris <Bryan_R_Harris [at] raytheon.com> writes:

BRH> Just as an academic exercise, I thought I should be able to do this:

BRH> **************************************
BRH> [at] a=(l=>35,k=>31,r=>7,k=>6);
BRH> [at] r=qw/l r r k/;

try to use some white space in your code. even in short examples it
helps.


BRH> # make an anonymous hash using [at] a, then grab values from it using
BRH> [at] r as keys

BRH> [at] a={ [at] a}{ [at] r};

learn the first rule of dereferencing. take a reference and wrap it in
{} with the proper sigil prefix. so to deref a hash, wrap it in %{}.

so you are making an anon hash around [at] a but not dereferencing it. you
need %{{ [at] a}}.

then to take a slice of that, you would change the % to [at] :

[at] {{ [at] a}}{ [at] r}

BRH> ... but it doesn't like the {}{} notation. Is this not possible?
BRH> I do this with lists all the time, e.g. [at] a = (split)[1,7,15];

that isn't the same. that is a slice of a list. you had an anon hash
which must be dereferenced first before you can slice it.

uri

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