--MrNobo1024
s]]HrLfbfe|EbBibmv]e|s}w}ciZx^RYhL}e^print

Nothing like an opportunity to improve:

sprintf'%*vX',$",@_

That's 19 characters by my count. Using the internal equivalent of space, plus why pop when you can just fire it in and hope for the best?

Am I correct in assuming the function has to return the string instead of printing it? One character either way.

I also think this only works in Perl 5.6 or better. A test in 5.005_02 fails, printing %*vX In that case, I think a 33 character solution is:

$_=unpack"H*",pop;s/..\B/$& /g;$_

As a note, I would have used @_ in the second one, but under an older version, it would seem that the unpack function is prototyped and this array is rendered to hex 31, which is in fact an ASCII "1". This could save one stroke, for a total of 32.

I do like jmcnamara's innovative solution.

Nice MrNobo. You can reach a compromise in length between your two approaches using:

printf '%*vX', ' ', pop;
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Wow, thanks guys. I learned something useful here.

I knew about %v, or thought I did: I didn't realize it was a modifier, or that I could change the dot to the space!

So (reading the perlfunc docs) what is the meaning of the V modifier? I don't understand the explaination "interpret integer as Perl's standard integer type"

—John

$_ = unpack("h*", shift); s/([0-9a-f]{2})/$1 /g; print;
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Simpler, if not much shorter ; )

Update: Hmm, this'll work too:

printf("%02x ", $_) for (unpack ("C*", shift));
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2nd Update: I think I'll be trading in my extra whitespaces for a 'chop;'

sub hexout
{
  #23456789012345678901234567890123 - 33 char
  $_=unpack"H*",pop;s/(..)/$1 /g;$_
}
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but it appends a space at the end

sub hexout
{
  #234567890123456789012 - 22
  sprintf'%*v2X',' ',pop
}
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#2 was with a little help from the sprintf docs

Update: Spent a little too much time. MrNobo1024 beat me to the punch ;-)

    sub hexout4  {
        "@{[unpack('H*',pop)=~/../g]}"
    }
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    printf "%#x\n", 0;
    printf "%#x\n", 1;

    __END__
    Prints:
    0
    0x1
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I guess that the behaviour comes from the C compiler that builds perl because the following gives the same results (at least where I could test it):

    int main () {

        printf("%#x\n", 0);
        printf("%#x\n", 1);

        return 0;
    }
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So why is 0 treated differently?

--
John.

Perl implements its own printf formatting. But if it follows the ANSI/ISO C specification for what %#X does, it is per spec: any non-zero output is prefixed with 0x.

To always have the prifix, just mention it in the string: "0x%X".

I like your idea of using interpolation instead of join.

—John