Re: register allocation: basic blocks, liveness and next use

Gene <gene.ressler@gmail.com>
Sat, 22 Mar 2008 20:47:01 -0700 (PDT)

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Related articles
register allocation: basic blocks, liveness and next use kevin.phillips83@yahoo.com (kphillips) (2008-03-22)
Re: register allocation: basic blocks, liveness and next use gene.ressler@gmail.com (Gene) (2008-03-22)
Re: register allocation: basic blocks, liveness and next use max@gustavus.edu (Max Hailperin) (2008-03-23)
Re: register allocation: basic blocks, liveness and next use max@gustavus.edu (Max Hailperin) (2008-03-23)
Re: register allocation: basic blocks, liveness and next use max@gustavus.edu (Max Hailperin) (2008-03-23)
Re: register allocation: basic blocks, liveness and next use kevin.phillips83@yahoo.com (kphillips) (2008-03-23)
Re: register allocation: basic blocks, liveness and next use gene.ressler@gmail.com (Gene) (2008-03-23)
Re: register allocation: basic blocks, liveness and next use cfc@shell01.TheWorld.com (Chris F Clark) (2008-03-23)
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From: Gene <gene.ressler@gmail.com>
Newsgroups: comp.compilers
Date: Sat, 22 Mar 2008 20:47:01 -0700 (PDT)
Organization: Compilers Central
References: 08-03-084
Keywords: registers
Posted-Date: 23 Mar 2008 10:28:11 EDT

On Mar 22, 8:22 pm, kphillips <kevin.phillip...@yahoo.com> wrote:
> The TAC would be something of this sort:
>
> == Block 1 =========
> PUSHPARAM 2
> CALL t1 A //call A and store the return value in
> t1
> == Block 2 =========
> PUSHPARAM -3
> CALL t2 A
> == Block 3 =========
> ADD t3 = t1, t2
> PRINT t3
> =================
>
> The problem is that the temporaries in both block 1 (t1) and block 2
> (t2) would be discarded (assumed dead since they are at the end of the
> block) and block 3 can't compute the addition.
>
> Am I missing something here?


Perhaps the definition of a basic block.


The definition of a temporary requires that none of its defs can reach
the end of the corresponding block. Obviously your t1..3 don't meet
this requirement.


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