Re: Implement Arithmetic Shift Right for signed number in perl

in reply to Implement Arithmetic Shift Right for signed number in perl

What is wrong with

printf '0x%x', -0xaaaaaaaa >> 32;
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It gives the expected result.

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Comment on Re: Implement Arithmetic Shift Right for signed number in perl Download Code

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Re^2: Implement Arithmetic Shift Right for signed number in perl by IamAwesom3 (Initiate) on Apr 23, 2013 at 15:15 UTC
That's not the correct way, It will not be true every time, this time you knew the result. I figured out the correct way `my $Rm = 0xAAAAAAAA; my $Rd = $Rm >> 31; # any shift amount from 0 top 31 for 32 bit data. $Rd \|= 0xffffffff << (32 - 31); #it can be (32 - any shift amount) $Rd = $Rd & 0xffffffff; #mask bits more than 32 printf "Rd = 0x%x\t Rb = %x\n",$Rd, $Rm;` [download] Output `Rd = 0xffffffff Rb = 0xaaaaaaaa` [download]	[reply] [d/l] [select]

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Re^2: Implement Arithmetic Shift Right for signed number in perl
by IamAwesom3 (Initiate) on Apr 23, 2013 at 15:15 UTC


my $Rm = 0xAAAAAAAA;
my $Rd = $Rm >> 31; # any shift amount from 0 top 31 for 32 bit data. 

$Rd |= 0xffffffff << (32 - 31); #it can be (32 - any shift amount)
$Rd = $Rd & 0xffffffff; #mask bits more than 32

printf "Rd = 0x%x\t Rb = %x\n",$Rd, $Rm;
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Rd = 0xffffffff  Rb = 0xaaaaaaaa
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[reply]
[d/l]
[select]

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