One solution I've seen elsewhere turned the float into an int by multiplying by a set power of ten. The precision was set by the spec, so they knew they only need 4 decimal places. Then they packed the int in network order (that's the N in your second example). When they unpacked it, they divided by the same power of ten.

If you want this to be really fast, you can cook up something simple in C (or even Inline::C) to do the same thing. Since you are using pack(), your Perl and C implementations will probably look the same so once you figure it out in Perl, it should be easy to translate to C.

There was one time I had to do this for several gigs of data, and I actually pulled out the assembly language to do it. The target was the Motorola 68k family, which had several nice instructions to flip these things around (flip the words, then flip the bytes in the words). That was a huge win even over the C code. I only got away with that because I knew exactly which microprocessors we had to use for the problem (ah, the joys of custom built hardware!)

Good luck!

Hmm. That's an interesting result. Can you give me the output of

print join(" ", map { sprintf "%#02x", $_ }
    unpack("C*",pack("L",0x12345678))), "\n";
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on your system? XS won't be possible, by the way, as this is intentionally a pure-perl implementation.

sure 0x78 0x56 0x34 0x12

I'm not sure why you crossed out your first answer. It is correct; the way to change between big and little endian packed IEEE floats is to use reverse:

    $ perl -le 'print unpack "H*", reverse pack "f", 440'
    43dc0000

    $ perl -le 'print unpack "H*",         pack "f", 440'
    0000dc43
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--
John.

For completeness' sake: if you want some background on this madness, read about Endianness.

Makeshifts last the longest.