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;

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Output 
Rd = 0xffffffff  Rb = 0xaaaaaaaa

[download]

In reply to Re^2: Implement Arithmetic Shift Right for signed number in perl by IamAwesom3
in thread Implement Arithmetic Shift Right for signed number in perl by IamAwesom3

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