>>>>> "BM" == Bob McConnell <r...@cbord.com> writes:
BM> From: Uri Guttman >>>>>>> "BM" == Bob McConnell <r...@cbord.com> writes: >> BM> From: Bryan R Harris >> >> >> >> I need to convert a number like this: -3205.0569059 >> >> ... into an 8-byte double (big and little endian), e.g. 4f 3e 52 BM> 00 2a BM> bc 93 >> >> d3 (I just made up those 8 byte values). >> >> >> >> Is this easy in perl? Are long and short ints easy as well? >> BM> The sprintf() family is your friend. >> >> that will only generate text (hex and other formats). he needs pack >> which does exactly what he wants. read perlpacktut for a tutorial on >> pack/unpack and then perlfunc -f pack for the reference on it. BM> That statement just confuses me. His initial value of -3205.0569059 is BM> also text. It is the human readable representation of the number, and is BM> not anything like what it looks like inside the computer. He just asked BM> for a different format for that text. Why is sprintf not a reasonable BM> way to do that? he said number. perl will automatically convert a string number to a float. but he wants access to the bytes in that float (hence his mentioning of endian). perl can store a float, integer or string in a scalar. but the float can only be used as a number, you can't get at its bytes. pack solves that problem. uri -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
