Re: Copy propagation optimization

On Aug 7, 5:20 pm, Fernando <prone...@gmail.com> wrote:
> On Aug 6, 3:45 am, rajeshkann...@aol.in wrote:
>
> > Please clarify my doubt regarding copy propagation optimization.
>
> > (snip)
>
> > gsiVarA = gsiVarB + gsiVarC;
>
> > gsiVarB = 1 ; // One of expression's operand
> > is
> > redefined
>
> > gsiVarD = gsiVarA ; // Statement-B


this is constant propagation, which is a subset of copy propagation.


> > As shown in the above snip, can the compiler use the result of
> > expression
> > (gsiVarB + gsiVarC) in Statement B instead of gsiVarA.
>
> > I expect the below result
>
> > (snip)
>
> > gsiVarA = gsiVarB + gsiVarC;
>
> > load R1, gsiVarB
> > load R2, gsiVarC
> > Add R1, R2
> > Store R1, R3
>
> > .... // gsiVarB is redefined.
> > ....
>
> > gsiVarD = gsiVarA ; // Statement-B
>
> > Store R1, gsiVarD // Result of expression is used.
>
> > (snip end)
>
> > Please clarify whether the compiler can perform the above optimization.


> the code you wrote looks fine to me. However, I think that is not
> really called copy propagation. You would have copy propagation if,
> for instance, you had:
>
> gsiVarC = ...
> gsiVarB = gsiVarA // this is a copy
> gsiVarD = gsiVarC + gsiVarB
>
> - The optimization would yield:
>
> gsiVarC = ...
> gsiVarD = gsiVarC + gsiVarA // the result of the copy is used instead.


this is copy propagation. The OP wanted to know whether this would be
implemented in a compiler by default. The answer is that it depends on
the compiler and the optimization level one opts for. GNU C implements
this at +O1.


http://gcc.gnu.org/onlinedocs/gcc-4.4.1/gcc/Optimize-Options.html#Optimize-Op
tions


regards
-kamal

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