Gimple Pass
Hi, I've written a gimple pass, which counts for the number of assignment statements in the c code. I've used the lval of the GIMPLE_MODIFY_STMT and checking for VAR_DECL condition. It is working But I want to know what are the TREE_CODEs for other remaing constructs viz declration stmt, conditions, count for constants and how to use them in the gimple pass. Can anybody help in this regard Thanking -- View this message in context: http://www.nabble.com/Gimple-Pass-tp24643577p24643577.html Sent from the gcc - Dev mailing list archive at Nabble.com.
Re: Gimple Pass
Hi Nathan
I looked into tree.def, I added a case in the new gimple pass to count
the number of integer variables in the c code. the following is the piece of
code.
Here, the first case is working, but I'm not getting the second one. can u
pls help in this regard.
How to trace thru the other codes & there usage
switch (TREE_CODE(stmt)){
case GIMPLE_MODIFY_STMT:
lval=GIMPLE_STMT_OPERAND(stmt,0);
rval=GIMPLE_STMT_OPERAND(stmt,1);
if(TREE_CODE(lval) == VAR_DECL) {
if(!DECL_ARTIFICIAL(lval)){
//print_generic_stmt(stderr,stmt,0);
numassigns++;
}
totalassigns++;
}
break;
case DECL:
if(TREE_CODE(stmt) == VAR_DECL){
if(TREE_CODE(TREE_TYPE(stmt))==INTEGER_TYPE){
intvar++;
}
}
break;
Nathan Froyd-2 wrote:
>
> On Fri, Jul 24, 2009 at 05:20:16AM -0700, pms wrote:
>> But I want to know what are the TREE_CODEs for other remaing constructs
>> viz declration stmt, conditions, count for constants and how to use them
>> in
>> the gimple pass. Can anybody help in this regard
>
> The names are defined in tree.def.
>
> -Nathan
>
>
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regarding optimization options in phase ordering
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
p...@shiva:~/Desktop/Compilers/GCC/build/test$ ls
1.c1.c.011t.ehopt1.c.038t.release_ssa
1.c.001t.tu1.c.012t.eh 1.c.123t.optimized
1.c.003t.original 1.c.013t.cfg 1.c.125t.blocks
1.c.004t.gimple1.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.forwprop21.c.086t.loop
1.c.003t.original 1.c.054t.alias1.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.lower1.c.058t.copyprop21.c.093t.sccp
1.c.011t.ehopt1.c.059t.mergephi21.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.ifcombine1.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.cplxlower1.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.forwprop11.c.072t.phicprop11.c.118t.forwprop4
1.c.029t.addressables11.c.073t.dse1 1.c.119t.phiopt3
1.c.030t.esra 1.c.074t.reassoc1 1.c.121t.copyrename4
1.c.031t.copyprop11.c.075t.dce2 1.c.122t.uncprop
1.c.032t.mergephi11.c.076t.forwprop31.c.123t.optimized
1.c.033t.cddce1 1.c.077t.phiopt2 1.c.124t.nrv
1.c.034t.sdse 1.c.078t.objsz1.c.125t.blocks
1.c.036t.switchconv 1.c.079t.ccp3 1.c.126t.final_cleanup
1.c.037t.profile 1.c.080t.copyprop31.c.203t.statistics
1.c.038t.release_ssa 1.c.081t.fab 1.s
1.c.048t.addressables21.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
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Re: regarding optimization options in phase ordering
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)
callprintf
pls reply
Richard Guenther-2 wrote:
>
> On Fri, Aug 7, 2009 at 1:50 PM, pms 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
>> p...@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.
>>
>>
>
>
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Re: regarding optimization options in phase ordering
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)
> callprintf
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)
> callprintf
The first four statements set up the stack. The last three do the
printf statement.
What is your real question?
Ian
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