* regarding optimization options in phase ordering
@ 2009-08-07 11:53 pms
2009-08-07 12:34 ` Richard Guenther
0 siblings, 1 reply; 6+ messages in thread
From: pms @ 2009-08-07 11:53 UTC (permalink / raw)
To: gcc
Hi,
We've a problem here. we were trying to use cc1 with & without -O
option to verify the optimizations happening in our sample code. these r the
list of outputs after each compilation
without -O
pms@Shiva:~/Desktop/Compilers/GCC/build/test$ ls
1.c 1.c.011t.ehopt 1.c.038t.release_ssa
1.c.001t.tu 1.c.012t.eh 1.c.123t.optimized
1.c.003t.original 1.c.013t.cfg 1.c.125t.blocks
1.c.004t.gimple 1.c.014t.cplxlower0 1.c.126t.final_cleanup
1.c.006t.vcg 1.c.015t.veclower 1.c.203t.statistics
1.c.007t.useless 1.c.021t.cleanup_cfg 1.s
1.c.010t.lower 1.c.023t.ssa a.out
with -O
1.c 1.c.051t.ccp2 1.c.085t.sink
1.c.001t.tu 1.c.052t.forwprop2 1.c.086t.loop
1.c.003t.original 1.c.054t.alias 1.c.087t.loopinit
1.c.004t.gimple 1.c.055t.retslot 1.c.088t.copyprop4
1.c.006t.vcg 1.c.056t.phiprop 1.c.089t.dceloop1
1.c.007t.useless 1.c.057t.fre 1.c.090t.lim
1.c.010t.lower 1.c.058t.copyprop2 1.c.093t.sccp
1.c.011t.ehopt 1.c.059t.mergephi2 1.c.094t.empty
1.c.012t.eh 1.c.061t.dce1 1.c.099t.ivcanon
1.c.013t.cfg 1.c.062t.cselim 1.c.104t.cunroll
1.c.015t.veclower 1.c.063t.ifcombine 1.c.107t.ivopts
1.c.021t.cleanup_cfg 1.c.064t.phiopt1 1.c.108t.loopdone
1.c.023t.ssa 1.c.066t.ch 1.c.111t.reassoc2
1.c.024t.early_optimizations 1.c.068t.cplxlower 1.c.113t.dom2
1.c.025t.einline2 1.c.069t.sra 1.c.114t.phicprop2
1.c.026t.copyrename1 1.c.070t.copyrename3 1.c.115t.cddce2
1.c.027t.ccp1 1.c.071t.dom1 1.c.117t.dse2
1.c.028t.forwprop1 1.c.072t.phicprop1 1.c.118t.forwprop4
1.c.029t.addressables1 1.c.073t.dse1 1.c.119t.phiopt3
1.c.030t.esra 1.c.074t.reassoc1 1.c.121t.copyrename4
1.c.031t.copyprop1 1.c.075t.dce2 1.c.122t.uncprop
1.c.032t.mergephi1 1.c.076t.forwprop3 1.c.123t.optimized
1.c.033t.cddce1 1.c.077t.phiopt2 1.c.124t.nrv
1.c.034t.sdse 1.c.078t.objsz 1.c.125t.blocks
1.c.036t.switchconv 1.c.079t.ccp3 1.c.126t.final_cleanup
1.c.037t.profile 1.c.080t.copyprop3 1.c.203t.statistics
1.c.038t.release_ssa 1.c.081t.fab 1.s
1.c.048t.addressables2 1.c.082t.sincos a.out
1.c.049t.copyrename2 1.c.083t.crited
But here, we tried to see the differences, until 1.c.027t.ccp1, the output
for the following source 1.c
#include
int main()
{
int a=5;
int b;
b=a;
printf("the number is :%d",b);
}
was 1.c.026t.copyrename1
;; Function main (main)
main ()
{
int b;
int a;
:
a_2 = b_1(D);
return;
}
but in 1.c.027t.ccp1, the output doesnot contain the actual assignment b=a.
;; Function main (main)
main ()
{
int b;
int a;
:
return;
}
We want to know, without b=a, how is it generating the final code for b=a
Kindly help
--
View this message in context: http://www.nabble.com/regarding-optimization-options-in-phase-ordering-tp24863416p24863416.html
Sent from the gcc - Dev mailing list archive at Nabble.com.
^ permalink raw reply [flat|nested] 6+ messages in thread* Re: regarding optimization options in phase ordering
2009-08-07 11:53 regarding optimization options in phase ordering pms
@ 2009-08-07 12:34 ` Richard Guenther
2009-08-07 13:11 ` pms
0 siblings, 1 reply; 6+ messages in thread
From: Richard Guenther @ 2009-08-07 12:34 UTC (permalink / raw)
To: pms; +Cc: gcc
On Fri, Aug 7, 2009 at 1:50 PM, pms<pmshiva@gmail.com> wrote:
>
> Hi,
> We've a problem here. we were trying to use cc1 with & without -O
> option to verify the optimizations happening in our sample code. these r the
> list of outputs after each compilation
> without -O
> pms@Shiva:~/Desktop/Compilers/GCC/build/test$ ls
> 1.c 1.c.011t.ehopt 1.c.038t.release_ssa
> 1.c.001t.tu 1.c.012t.eh 1.c.123t.optimized
> 1.c.003t.original 1.c.013t.cfg 1.c.125t.blocks
> 1.c.004t.gimple 1.c.014t.cplxlower0 1.c.126t.final_cleanup
> 1.c.006t.vcg 1.c.015t.veclower 1.c.203t.statistics
> 1.c.007t.useless 1.c.021t.cleanup_cfg 1.s
> 1.c.010t.lower 1.c.023t.ssa a.out
>
> with -O
>
> 1.c 1.c.051t.ccp2 1.c.085t.sink
> 1.c.001t.tu 1.c.052t.forwprop2 1.c.086t.loop
> 1.c.003t.original 1.c.054t.alias 1.c.087t.loopinit
> 1.c.004t.gimple 1.c.055t.retslot 1.c.088t.copyprop4
> 1.c.006t.vcg 1.c.056t.phiprop 1.c.089t.dceloop1
> 1.c.007t.useless 1.c.057t.fre 1.c.090t.lim
> 1.c.010t.lower 1.c.058t.copyprop2 1.c.093t.sccp
> 1.c.011t.ehopt 1.c.059t.mergephi2 1.c.094t.empty
> 1.c.012t.eh 1.c.061t.dce1 1.c.099t.ivcanon
> 1.c.013t.cfg 1.c.062t.cselim 1.c.104t.cunroll
> 1.c.015t.veclower 1.c.063t.ifcombine 1.c.107t.ivopts
> 1.c.021t.cleanup_cfg 1.c.064t.phiopt1 1.c.108t.loopdone
> 1.c.023t.ssa 1.c.066t.ch 1.c.111t.reassoc2
> 1.c.024t.early_optimizations 1.c.068t.cplxlower 1.c.113t.dom2
> 1.c.025t.einline2 1.c.069t.sra 1.c.114t.phicprop2
> 1.c.026t.copyrename1 1.c.070t.copyrename3 1.c.115t.cddce2
> 1.c.027t.ccp1 1.c.071t.dom1 1.c.117t.dse2
> 1.c.028t.forwprop1 1.c.072t.phicprop1 1.c.118t.forwprop4
> 1.c.029t.addressables1 1.c.073t.dse1 1.c.119t.phiopt3
> 1.c.030t.esra 1.c.074t.reassoc1 1.c.121t.copyrename4
> 1.c.031t.copyprop1 1.c.075t.dce2 1.c.122t.uncprop
> 1.c.032t.mergephi1 1.c.076t.forwprop3 1.c.123t.optimized
> 1.c.033t.cddce1 1.c.077t.phiopt2 1.c.124t.nrv
> 1.c.034t.sdse 1.c.078t.objsz 1.c.125t.blocks
> 1.c.036t.switchconv 1.c.079t.ccp3 1.c.126t.final_cleanup
> 1.c.037t.profile 1.c.080t.copyprop3 1.c.203t.statistics
> 1.c.038t.release_ssa 1.c.081t.fab 1.s
> 1.c.048t.addressables2 1.c.082t.sincos a.out
> 1.c.049t.copyrename2 1.c.083t.crited
>
> But here, we tried to see the differences, until 1.c.027t.ccp1, the output
> for the following source 1.c
> #include
> int main()
>
> {
> int a=5;
> int b;
> b=a;
> printf("the number is :%d",b);
> }
>
> was 1.c.026t.copyrename1
> ;; Function main (main)
>
> main ()
> {
> int b;
> int a;
>
> :
> a_2 = b_1(D);
> return;
>
> }
> but in 1.c.027t.ccp1, the output doesnot contain the actual assignment b=a.
> ;; Function main (main)
>
> main ()
> {
> int b;
> int a;
>
> :
> return;
>
> }
>
> We want to know, without b=a, how is it generating the final code for b=a
Nothing. Because it's a dead statement.
Richard.
> Kindly help
>
> --
> View this message in context: http://www.nabble.com/regarding-optimization-options-in-phase-ordering-tp24863416p24863416.html
> Sent from the gcc - Dev mailing list archive at Nabble.com.
>
>
^ permalink raw reply [flat|nested] 6+ messages in thread* Re: regarding optimization options in phase ordering
2009-08-07 12:34 ` Richard Guenther
@ 2009-08-07 13:11 ` pms
2009-08-10 11:20 ` Ian Lance Taylor
0 siblings, 1 reply; 6+ messages in thread
From: pms @ 2009-08-07 13:11 UTC (permalink / raw)
To: gcc
thanks, But b=a is a assignment statement. It is doing some memory operations
isn't it. Assuming b=a is a dead statement, how r the following i386
assembly statements generated
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $16, %esp
movl $5, 4(%esp)
movl $.LC0, (%esp)
call printf
pls reply
Richard Guenther-2 wrote:
>
> On Fri, Aug 7, 2009 at 1:50 PM, pms<pmshiva@gmail.com> wrote:
>>
>> Hi,
>> We've a problem here. we were trying to use cc1 with & without -O
>> option to verify the optimizations happening in our sample code. these r
>> the
>> list of outputs after each compilation
>> without -O
>> pms@Shiva:~/Desktop/Compilers/GCC/build/test$ ls
>> 1.c 1.c.011t.ehopt 1.c.038t.release_ssa
>> 1.c.001t.tu 1.c.012t.eh 1.c.123t.optimized
>> 1.c.003t.original 1.c.013t.cfg 1.c.125t.blocks
>> 1.c.004t.gimple 1.c.014t.cplxlower0 1.c.126t.final_cleanup
>> 1.c.006t.vcg 1.c.015t.veclower 1.c.203t.statistics
>> 1.c.007t.useless 1.c.021t.cleanup_cfg 1.s
>> 1.c.010t.lower 1.c.023t.ssa a.out
>>
>> with -O
>>
>> 1.c 1.c.051t.ccp2 1.c.085t.sink
>> 1.c.001t.tu 1.c.052t.forwprop2 1.c.086t.loop
>> 1.c.003t.original 1.c.054t.alias 1.c.087t.loopinit
>> 1.c.004t.gimple 1.c.055t.retslot 1.c.088t.copyprop4
>> 1.c.006t.vcg 1.c.056t.phiprop 1.c.089t.dceloop1
>> 1.c.007t.useless 1.c.057t.fre 1.c.090t.lim
>> 1.c.010t.lower 1.c.058t.copyprop2 1.c.093t.sccp
>> 1.c.011t.ehopt 1.c.059t.mergephi2 1.c.094t.empty
>> 1.c.012t.eh 1.c.061t.dce1 1.c.099t.ivcanon
>> 1.c.013t.cfg 1.c.062t.cselim 1.c.104t.cunroll
>> 1.c.015t.veclower 1.c.063t.ifcombine 1.c.107t.ivopts
>> 1.c.021t.cleanup_cfg 1.c.064t.phiopt1 1.c.108t.loopdone
>> 1.c.023t.ssa 1.c.066t.ch 1.c.111t.reassoc2
>> 1.c.024t.early_optimizations 1.c.068t.cplxlower 1.c.113t.dom2
>> 1.c.025t.einline2 1.c.069t.sra 1.c.114t.phicprop2
>> 1.c.026t.copyrename1 1.c.070t.copyrename3 1.c.115t.cddce2
>> 1.c.027t.ccp1 1.c.071t.dom1 1.c.117t.dse2
>> 1.c.028t.forwprop1 1.c.072t.phicprop1 1.c.118t.forwprop4
>> 1.c.029t.addressables1 1.c.073t.dse1 1.c.119t.phiopt3
>> 1.c.030t.esra 1.c.074t.reassoc1 1.c.121t.copyrename4
>> 1.c.031t.copyprop1 1.c.075t.dce2 1.c.122t.uncprop
>> 1.c.032t.mergephi1 1.c.076t.forwprop3 1.c.123t.optimized
>> 1.c.033t.cddce1 1.c.077t.phiopt2 1.c.124t.nrv
>> 1.c.034t.sdse 1.c.078t.objsz 1.c.125t.blocks
>> 1.c.036t.switchconv 1.c.079t.ccp3
>> 1.c.126t.final_cleanup
>> 1.c.037t.profile 1.c.080t.copyprop3 1.c.203t.statistics
>> 1.c.038t.release_ssa 1.c.081t.fab 1.s
>> 1.c.048t.addressables2 1.c.082t.sincos a.out
>> 1.c.049t.copyrename2 1.c.083t.crited
>>
>> But here, we tried to see the differences, until 1.c.027t.ccp1, the
>> output
>> for the following source 1.c
>> #include
>> int main()
>>
>> {
>> int a=5;
>> int b;
>> b=a;
>> printf("the number is :%d",b);
>> }
>>
>> was 1.c.026t.copyrename1
>> ;; Function main (main)
>>
>> main ()
>> {
>> int b;
>> int a;
>>
>> :
>> a_2 = b_1(D);
>> return;
>>
>> }
>> but in 1.c.027t.ccp1, the output doesnot contain the actual assignment
>> b=a.
>> ;; Function main (main)
>>
>> main ()
>> {
>> int b;
>> int a;
>>
>> :
>> return;
>>
>> }
>>
>> We want to know, without b=a, how is it generating the final code for b=a
>
> Nothing. Because it's a dead statement.
>
> Richard.
>
>> Kindly help
>>
>> --
>> View this message in context:
>> http://www.nabble.com/regarding-optimization-options-in-phase-ordering-tp24863416p24863416.html
>> Sent from the gcc - Dev mailing list archive at Nabble.com.
>>
>>
>
>
--
View this message in context: http://www.nabble.com/regarding-optimization-options-in-phase-ordering-tp24863416p24864601.html
Sent from the gcc - Dev mailing list archive at Nabble.com.
^ permalink raw reply [flat|nested] 6+ messages in thread* Re: regarding optimization options in phase ordering
2009-08-07 13:11 ` pms
@ 2009-08-10 11:20 ` Ian Lance Taylor
2009-08-10 13:00 ` pms
0 siblings, 1 reply; 6+ messages in thread
From: Ian Lance Taylor @ 2009-08-10 11:20 UTC (permalink / raw)
To: pms; +Cc: gcc
pms <pmshiva@gmail.com> writes:
> thanks, But b=a is a assignment statement. It is doing some memory operations
> isn't it. Assuming b=a is a dead statement, how r the following i386
> assembly statements generated
> pushl %ebp
> movl %esp, %ebp
> andl $-16, %esp
> subl $16, %esp
> movl $5, 4(%esp)
> movl $.LC0, (%esp)
> call printf
The first four statements set up the stack. The last three do the
printf statement.
What is your real question?
Ian
^ permalink raw reply [flat|nested] 6+ messages in thread
* Re: regarding optimization options in phase ordering
2009-08-10 11:20 ` Ian Lance Taylor
@ 2009-08-10 13:00 ` pms
2009-08-10 13:08 ` Ian Lance Taylor
0 siblings, 1 reply; 6+ messages in thread
From: pms @ 2009-08-10 13:00 UTC (permalink / raw)
To: gcc
Thanx Ian,
My question is as follwos
We've a problem here. we were trying to use cc1 with & without -O option to
verify the optimizations happening in our sample code.
the sample code is given below
file name : 1.c
#include
int main()
{ int a=5;
int b;
b=a;
printf("the number is :%d",b); }
Here, in 1.c.026t.copyrename1, we get the following output
1.c.026t.copyrename1
;; Function main (main)
main ()
{ int b; int a; : a_2 = b_1(D); return; }
but in 1.c.027t.ccp1, the output does not contain the actual assignment
b=a.
;; Function main (main)
main ()
{ int b; int a; : return; }
We want to know, without b=a, how is it generating the following final code
for b=a
pushl %ebp
> movl %esp, %ebp
> andl $-16, %esp
> subl $16, %esp
> movl $5, 4(%esp)
> movl $.LC0, (%esp)
> call printf
Kindly help
> thanks, But b=a is a assignment statement. It is doing some memory
> operations
> isn't it. Assuming b=a is a dead statement, how r the following i386
> assembly statements generated
> pushl %ebp
> movl %esp, %ebp
> andl $-16, %esp
> subl $16, %esp
> movl $5, 4(%esp)
> movl $.LC0, (%esp)
> call printf
The first four statements set up the stack. The last three do the
printf statement.
What is your real question?
Ian
--
View this message in context: http://www.nabble.com/regarding-optimization-options-in-phase-ordering-tp24863416p24898021.html
Sent from the gcc - Dev mailing list archive at Nabble.com.
^ permalink raw reply [flat|nested] 6+ messages in thread* Re: regarding optimization options in phase ordering
2009-08-10 13:00 ` pms
@ 2009-08-10 13:08 ` Ian Lance Taylor
0 siblings, 0 replies; 6+ messages in thread
From: Ian Lance Taylor @ 2009-08-10 13:08 UTC (permalink / raw)
To: pms; +Cc: gcc
pms <pmshiva@gmail.com> writes:
> My question is as follwos
>
> We've a problem here. we were trying to use cc1 with & without -O option to
> verify the optimizations happening in our sample code.
> the sample code is given below
> file name : 1.c
> #include
> int main()
> { int a=5;
> int b;
> b=a;
> printf("the number is :%d",b); }
> Here, in 1.c.026t.copyrename1, we get the following output
>
> 1.c.026t.copyrename1
> ;; Function main (main)
> main ()
> { int b; int a; : a_2 = b_1(D); return; }
> but in 1.c.027t.ccp1, the output does not contain the actual assignment
> b=a.
> ;; Function main (main)
> main ()
> { int b; int a; : return; }
> We want to know, without b=a, how is it generating the following final code
> for b=a
> pushl %ebp
>> movl %esp, %ebp
>> andl $-16, %esp
>> subl $16, %esp
>> movl $5, 4(%esp)
>> movl $.LC0, (%esp)
>> call printf
No code is generated for the statement "b=a". In effect, the program is
optimized into
printf("the number is :%d",5);
as that is equivalent to the original program.
This is a very simple compiler optimization. Please consider taking
future questions about code generation to the mailing list
gcc-help@gcc.gnu.org. The mailing list gcc@gcc.gnu.org is for gcc
developers. Thanks.
Ian
>> thanks, But b=a is a assignment statement. It is doing some memory
>> operations
>> isn't it. Assuming b=a is a dead statement, how r the following i386
>> assembly statements generated
>> pushl %ebp
>> movl %esp, %ebp
>> andl $-16, %esp
>> subl $16, %esp
>> movl $5, 4(%esp)
>> movl $.LC0, (%esp)
>> call printf
>
> The first four statements set up the stack. The last three do the
> printf statement.
>
> What is your real question?
>
> Ian
^ permalink raw reply [flat|nested] 6+ messages in thread
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2009-08-07 11:53 regarding optimization options in phase ordering pms
2009-08-07 12:34 ` Richard Guenther
2009-08-07 13:11 ` pms
2009-08-10 11:20 ` Ian Lance Taylor
2009-08-10 13:00 ` pms
2009-08-10 13:08 ` Ian Lance Taylor
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