IamAwesom3 has asked for the wisdom of the Perl Monks concerning the following question:
Hello monks,
I am trying to implement Arithmetic Shift Right for signed number in perl.
This is how it works.
$Ra = "0xAAAAAAAA"; then if I perform ASR 32 times the final value should be 0xFFFFFFFF. It copies sign bit from bit 31 to bit 31 and also to bit 30 and bit 30's data to bit 29 and so on.
I read it online that if I use "use integer" in same scope as the shift operand then It will perform ASR.
I wrote a code like this
Output ismy $Rm = "0xAAAAAAAA"; use integer; my $Rd = hex($Rm) >> 31; printf "Rd = 0x%x\t Rb = 0x%x\n",$Rd, hex($Rm);
I am expecting Rd to be 0xFFFFFFFF but it doesn't seems to be working.Rd = 0x1 Rb = 0xaaaaaaaa
Can someone please help me?
Thanks P
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Replies are listed 'Best First'. | |
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Re: Implement Arithmetic Shift Right for signed number in perl
by Krambambuli (Curate) on Apr 23, 2013 at 08:36 UTC | |
Re: Implement Arithmetic Shift Right for signed number in perl
by choroba (Cardinal) on Apr 22, 2013 at 23:59 UTC | |
by IamAwesom3 (Initiate) on Apr 23, 2013 at 15:15 UTC |
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