perl 操作位串和二进制文件 问题

picbhan

perl中如何把两个数转换成bit后连接起来呢？
如我有两个数， 31 和 37294798， 想把 31 转换成长度为5的bit，然后用27位来储存 37294798， 最后连接成一个32位的vec，有没有好的建议？

另外，如果我的一个str 是由vec生成的很多个这样以32位为单元的二进制串，如何写入到一个binary文件中呢？再如何读出来？因为长度不固定，但是可以一个str写成一行，然后整行读取分析。

最近在研究pack， unpack， vec等函数，但是一直没吃透，希望有懂的可以指点下。

dugu072

其实，你的需求用sprintf、oct等来做，更容易理解吧，但性能倒是不确认～
我对pack那些模板参数倒是也不熟，翻了下perldoc，纯粹用pack、unpack实现你的需求：

unpack( I, pack( 'b*', unpack( 'b27', pack('i', 37294798 ) ) . unpack( 'b5', pack('c',31) ) ) )

返回的是unsigned，然后你上面说的 5 bits + 27 bits，没说清楚是高、低位，上面的代码把5bits放在了 高位。

jason680

回复 1# picbhan

It seems that can't use vec for your question because the BITS must be 1,2,4,8,16...

$ perldoc -f vec
       vec EXPR,OFFSET,BITS
               Treats the string in EXPR as a bit vector made up of elements
               of width BITS, and returns the value of the element specified
               by OFFSET as an unsigned integer.  BITS therefore specifies the
               number of bits that are reserved for each element in the bit
               vector.  This must be a power of two from 1 to 32 (or 64, if
               your platform supports that).

               If BITS is 8, "elements" coincide with bytes of the input
               string.

               If BITS is 16 or more, bytes of the input string are grouped
               into chunks of size BITS/8, and each group is converted to a
               number as with pack()/unpack() with big-endian formats "n"/"N"
               (and analogously for BITS==64).  See "pack" for details.


   

picbhan

回复 2# dugu072


    这两天没时间看论坛，没回复，不好意思。

sprintf 加 vec 加 pack / unpack 可以做到， 单纯用vec 也可以。
$c = sprintf("%05b", $a) . sprintf("%027b", $b);
vec($d, 0, 32) = unpack("N", pack("B32", $c));


只用vec需要做位移操作和或操作。
$a = '';
vec($a, 0, 32) = 31;
vec($b, 0, 32) = $d << 5;
$a = $a | $b;

picbhan

回复 3# jason680


    please see #4.

jason680

回复 5# picbhan

How about this easy way to do it as below:

$value32 =  ($a_5bits << 27 ) + $b_27bits;

or you want it must be 5 and 27 bits only

$value32 =  (($a_5bits & 0x11)<< 27 ) + ($b_27bits & 0x7FFFFFF);


$a_5bits = $value32 >> 27;
$b_27bits = ($value32  & 0x7FFFFFF);  

picbhan

回复 6# jason680


    Actually, it's not exactly 5 bits and 27 bits.
I use the last 5 bits to store values between 0 and 31, and use first 28 bits to store a very big value (it at most needs 28 bits), hence the last bit of these 28 bits is removed and overlaped by the first bit of the 5 bits, when deparse this 32 bits value, I first deparse last 5 bits to get the small value between 0 and 31, and the add a 0 to the first 27 bits to get the big value, which may 1 less than its orignal value, but it's OK.

Hence I can't sure the easy way you provided will work fine or not.

Like this:
$value32 = (($b_27bits >> 1) << 5) + $a_5bits;

I think this will work fine.

Anyway, thank you!

sjdy521

回复 7# picbhan


    他跟你说英文，你就不必也用英文回他了吧。。

picbhan

回复 8# sjdy521


    无所谓的，我们大多时候也是用英文写邮件什么的，呵呵

cdtits

学习学习，也许以后能用