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Re: Easier way to load a buffer?

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From:
ToddAndMargo via perl6-users
Date:
June 10, 2022 20:48
Subject:
Re: Easier way to load a buffer?
Message ID:
7eaa423e-4791-df12-1868-cd58ad189dba@zoho.com
>> On Fri, 10 Jun 2022 at 15:49, ToddAndMargo via perl6-users 
>> <perl6-users@perl.org <mailto:perl6-users@perl.org>> wrote:
>> 
>>     Hi All,
>> 
>>     I am looking for an easier way to load a buffer.
>> 
>>     I know about this way
>> 
>>     [4] > my Buf $b=Buf.new(0x2A, 0x54, 0xFF, 0x53);
>>     Buf:0x<2A 54 FF 53>
>> 
>>     I would like to do it on one big blast:
>> 
>>     my Buf $b=Buf.new(0x2A54FF53A5F1D36F1CEA7E61FC37A20D54A77FE7B78);
>>     Cannot unbox 170 bit wide bigint into native integer
>> 
>>     But do not know the proper syntax.
>> 
>>     Any words of wisdom?  Am I stuck with the hard way?
>> 
>>     Many thanks,
>>     -T
>> 

On 6/10/22 08:36, Simon Proctor wrote:
> So Buf is expecting a list of integers. If' you've got one long one in a 
> string like "2A54FF53A5F1D36F1CEA7E61FC37A20D54A77FE7B78" you want to 
> split it into smaller values so something like this?
> 
> my $hex = "2A54FF53A5F1D36F1CEA7E61FC37A20D54A77FE7B78";
> my Buf $b=Buf.new(comb(2).map( *.parse-base(16) ));
> 
> Which does the trick I think.

What does the * i *.parse-base do?

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