如何在Perl中解压缩双精度值？

Question

From this question:

bytearray - Perl pack/unpack and length of binary string - Stack Overflow

I've learned that @unparray = unpack("d "x5, $aa); in the snippet below results with string items in the unparray - not with double precision numbers (as I expected).

Is it possible to somehow obtain an array of double-precision values from the $aa bytestring in the snippet below?:

$a = pack("d",255);
print length($a)."\n";
# prints 8

$aa = pack("ddddd", 255,123,0,45,123);
print length($aa)."\n";
# prints 40

@unparray = unpack("d "x5, $aa);
print scalar(@unparray)."\n";
# prints 5

print length($unparray[0])."\n" 
# prints 3

printf "%d\n", $unparray[0] '
# prints 255

# one liner:
# perl -e '$a = pack("d",255); print length($a)."\n"; $aa = pack("ddddd", 255,123,0,45,123); print length($aa)."\n"; @unparray = unpack("d "x5, $aa); print scalar(@unparray)."\n"; print length($unparray[0])."\n" '

Many thanks in advance for any answers,
Cheers!

Answer 1

What makes you think it's not stored as a double?

use feature qw( say );

use Config      qw( %Config );
use Devel::Peek qw( Dump );

my @a = unpack "d5", pack "d5", 255,123,0,45,123;

say 0+@a;             # 5
Dump $a[0];           # NOK (floating point format)
say $Config{nvsize};  # 8 byte floats on this build

Answer 2

Sorry, but you've misunderstood hobbs' answer to your earlier question.

$unparray[0] is a double-precision floating-point value; but length is not like (say) C's sizeof operator, and doesn't tell you the size of its argument. Rather, it converts its argument to a string, and then tells you the length of that string.

For example, this:

my $a = 3.0 / 1.5;
print length($a), "\n";

will print this:

1

because it sets $a to 2.0 , which gets stringified as 2 , which has length 1 .