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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

    my $Rm = "0xAAAAAAAA";
    use integer;
    my $Rd = hex($Rm) >> 31;
    printf "Rd = 0x%x\t Rb = 0x%x\n",$Rd, hex($Rm);
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Output is

Rd = 0x1         Rb = 0xaaaaaaaa
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I am expecting Rd to be 0xFFFFFFFF but it doesn't seems to be working.

Can someone please help me?

Thanks P

In reply to Implement Arithmetic Shift Right for signed number in perl by IamAwesom3

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