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Posted to commits@harmony.apache.org by "George Timoshenko (JIRA)" <ji...@apache.org> on 2006/12/12 12:28:25 UTC

[jira] Created: (HARMONY-2633) [drlvm][jit][opt] simplifyTauDivOfMul does nothing in Simplifier

[drlvm][jit][opt] simplifyTauDivOfMul does nothing in Simplifier
----------------------------------------------------------------

                 Key: HARMONY-2633
                 URL: http://issues.apache.org/jira/browse/HARMONY-2633
             Project: Harmony
          Issue Type: Improvement
            Reporter: George Timoshenko
            Priority: Minor


Opnd*
Simplifier::simplifyTauDivOfMul(Modifier mod,
                                Type* type,
                                Opnd* src1,
                                Opnd* src2,
                                Opnd* tauCheckedOpnds) {
    //
    // if (c1%c2==0), then
    //   (c1 * s1) / c2 -> (c1/c2) * s1
    //
    return NULL;

}

IMO the idea from comment should be implemented. (c1 and c2 must be integers, and c1*s1 must not produce an overflow)


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[jira] Updated: (HARMONY-2633) [drlvm][jit][opt] simplifyTauDivOfMul does nothing in Simplifier

Posted by "George Timoshenko (JIRA)" <ji...@apache.org>.
     [ http://issues.apache.org/jira/browse/HARMONY-2633?page=all ]

George Timoshenko updated HARMONY-2633:
---------------------------------------

    Component/s: DRLVM

> [drlvm][jit][opt] simplifyTauDivOfMul does nothing in Simplifier
> ----------------------------------------------------------------
>
>                 Key: HARMONY-2633
>                 URL: http://issues.apache.org/jira/browse/HARMONY-2633
>             Project: Harmony
>          Issue Type: Improvement
>          Components: DRLVM
>            Reporter: George Timoshenko
>            Priority: Minor
>
> Opnd*
> Simplifier::simplifyTauDivOfMul(Modifier mod,
>                                 Type* type,
>                                 Opnd* src1,
>                                 Opnd* src2,
>                                 Opnd* tauCheckedOpnds) {
>     //
>     // if (c1%c2==0), then
>     //   (c1 * s1) / c2 -> (c1/c2) * s1
>     //
>     return NULL;
> }
> IMO the idea from comment should be implemented. (c1 and c2 must be integers, and c1*s1 must not produce an overflow)

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