search for: l2_globalpool1

Thanks Chris! That was causing me some confusion. Now it's producing a more meaningful output: poolinit((&l2_GlobalPool), 4u, 4u); poolinit((&l2_GlobalPool1), 4u, 4u); l5_tmp_2E_0_2E_i = poolalloc((&l2_GlobalPool), 4u); l6_tmp_2E_7_2E_i2 = l1_makeList((&l2_GlobalPool1), 0); *(&((struct l_struct_2E_list *)l5_tmp_2E_0_2E_i)->field0) = l6_tmp_2E_7_2E_i2; However, if I am understanding correctly, the first node is being allocated in o...

On Sat, 2 Apr 2005, Ricardo wrote: > After applying the PA to it, the output is something like this: ... > My question is: why is this malloc necessary? > ltmp_2_5 = malloc(4u); > Shouldn't be the result a program with this malloc replaced by poolalloc? > Should I include a special flag to achieve this? Ah, sorry, my memory was wrong. The default is to perform the

Thanks for the answer I am trying to test the PA with a program very similar to the one used as an example in the paper located here: http://llvm.cs.uiuc.edu/pubs/2003-04-29-DataStructureAnalysisTR.html The program is as follows: ========================== struct list { struct list *Next; }; struct list *makeList (int Num) { struct list *New = malloc ( sizeof ( struct list ) );