Hi,
I'm having a problem with the use iterator. Each "use" that I see,
when
using the use_iterator, is the same as the "def". Meaning, in the code
below the pDef is always equal to pUse pointer for every instruction in all
basic blocks (except terminators).
for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i)
Instruction* pDef = &(*i);
errs() << "Def: " << *pDef <<
"\n";
for (auto ui = pDef->use_begin(), uie = pDef->use_end();
ui
!= uie; ++ui)
{
Instruction* pUse = dyn_cast<Instruction>(*ui);
errs() << " Use: \t" << *pUse
<< "\n";
}
}
However, everything works as expected when using the range-based use
iterator with the following code.
for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i)
{
Instruction* pDef = &(*i);
errs() << "Def: " << *pDef <<
"\n";
for (User* pUser : pDef->users())
{
Instruction* pUse = dyn_cast<Instruction>(pUser);
errs() << " Use: \t" << *pUse
<< "\n";
}
}
Also, the code is executed inside a function pass. So was initially
thinking I somehow screwed up the use information in a previous pass.
However, I would assume the range-based iterator would not work as well but
it does.
Finally, I'm currently using LLVM 3.5.1 built for Windows. Google hasn't
been much help. Anybody have any suggestions as to why the first example
above doesn't work?
Thanks,
Zack
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On Tue, Jun 9, 2015 at 7:02 PM, Zack Waters <zswaters at gmail.com> wrote:> Hi, > > I'm having a problem with the use iterator. Each "use" that I see, when > using the use_iterator, is the same as the "def". Meaning, in the code below > the pDef is always equal to pUse pointer for every instruction in all basic > blocks (except terminators). > > for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i) > > Instruction* pDef = &(*i); > errs() << "Def: " << *pDef << "\n"; > > for (auto ui = pDef->use_begin(), uie = pDef->use_end(); ui > != uie; ++ui) > { > Instruction* pUse = dyn_cast<Instruction>(*ui);This doesn't do what you think it does. This gives you the places the instruction i occurs in some other instruction. Those places are in fact, pointers to i already. You are then calling operator *, which will return i, and then dyn_casting it to an instruction, which is still i. If you want the *instruction that uses i* (not the pointer to the place in that instruction that *is* i), then you want the users, not the uses. IE this would be Instruction* pUse = dyn_cast<Instruction>(ui->getUser()) Or just iterator over the users, not the uses.> errs() << " Use: \t" << *pUse << "\n"; > } > } > > However, everything works as expected when using the range-based use > iterator with the following code. > > for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i) > { > Instruction* pDef = &(*i); > errs() << "Def: " << *pDef << "\n"; > > for (User* pUser : pDef->users()) > {This is not a use iterator, it is a user iterator, which is giving you *the instructions that contain the uses*. To illustrate the difference. given a = foo b = a + c (I'm simplifying, since LLVM does not represent LHS'en or have copies) if you ask for a use iterator of a, you will get &(operand 1 of b, which is a pointer to a). If you dereference it, you will get "a" (and if you dyn_cast that to an instruction, it will give you a = foo) if you ask for it's user, you will get "a + c" If you ask for a user iterator of a you will get a + c> Instruction* pUse = dyn_cast<Instruction>(pUser); > errs() << " Use: \t" << *pUse << "\n"; > } > } > > Also, the code is executed inside a function pass. So was initially thinking > I somehow screwed up the use information in a previous pass. However, I > would assume the range-based iterator would not work as well but it does. >Because it is not, in fact, giving you uses :)
On 6/9/15 8:02 PM, Zack Waters wrote:> Hi, > > I'm having a problem with the use iterator. Each "use" that I see, when > using the use_iterator, is the same as the "def". Meaning, in the code > below the pDef is always equal to pUse pointer for every instruction in > all basic blocks (except terminators). > > for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i) > Instruction* pDef = &(*i); > errs() << "Def: " << *pDef << "\n"; > > for (auto ui = pDef->use_begin(), uie > pDef->use_end(); ui != uie; ++ui) > {'user' != 'use'. Think of llvm::Use as the edge between the place where a value is produced, and the place where that value is consumed. The consumer is the 'User', and the Use points at it. http://llvm.org/docs/doxygen/html/classllvm_1_1Use.html The confusing thing that's happening below is that the llvm::Use is implicitly converted via `llvm::Use::operator Value *() const` to a `Value*`, and that `Value*` is `pDef`. HTH, Jon> Instruction* pUse = dyn_cast<Instruction>(*ui); > errs() << " Use: \t" << *pUse << "\n"; > } > } > > However, everything works as expected when using the range-based use > iterator with the following code. > > for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i) > { > Instruction* pDef = &(*i); > errs() << "Def: " << *pDef << "\n"; > > for (User* pUser : pDef->users()) > { > Instruction* pUse = dyn_cast<Instruction>(pUser); > errs() << " Use: \t" << *pUse << "\n"; > } > } > > Also, the code is executed inside a function pass. So was initially > thinking I somehow screwed up the use information in a previous pass. > However, I would assume the range-based iterator would not work as well > but it does. > > Finally, I'm currently using LLVM 3.5.1 built for Windows. Google hasn't > been much help. Anybody have any suggestions as to why the first example > above doesn't work? > > Thanks, > Zack > > > _______________________________________________ > LLVM Developers mailing list > LLVMdev at cs.uiuc.edu http://llvm.cs.uiuc.edu > http://lists.cs.uiuc.edu/mailman/listinfo/llvmdev >-- Jon Roelofs jonathan at codesourcery.com CodeSourcery / Mentor Embedded
Thanks Dan and Jon. I made an incorrect assumption that the "use" iterator was actually giving me the "user" when de-referencing it. Did it always have this behavior in previous LLVM versions? I've seen lots of examples of the "use" iterator being dereferenced and resulting Instruction pointer being treated as the "user"? Thanks, Zack On Tue, Jun 9, 2015 at 7:25 PM, Jonathan Roelofs <jonathan at codesourcery.com> wrote:> > > On 6/9/15 8:02 PM, Zack Waters wrote: > >> Hi, >> >> I'm having a problem with the use iterator. Each "use" that I see, when >> using the use_iterator, is the same as the "def". Meaning, in the code >> below the pDef is always equal to pUse pointer for every instruction in >> all basic blocks (except terminators). >> >> for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i) >> Instruction* pDef = &(*i); >> errs() << "Def: " << *pDef << "\n"; >> >> for (auto ui = pDef->use_begin(), uie >> pDef->use_end(); ui != uie; ++ui) >> { >> > > 'user' != 'use'. > > Think of llvm::Use as the edge between the place where a value is > produced, and the place where that value is consumed. The consumer is the > 'User', and the Use points at it. > > http://llvm.org/docs/doxygen/html/classllvm_1_1Use.html > > The confusing thing that's happening below is that the llvm::Use is > implicitly converted via `llvm::Use::operator Value *() const` to a > `Value*`, and that `Value*` is `pDef`. > > > HTH, > > Jon > > Instruction* pUse = dyn_cast<Instruction>(*ui); >> errs() << " Use: \t" << *pUse << "\n"; >> } >> } >> >> However, everything works as expected when using the range-based use >> iterator with the following code. >> >> for (auto i = inst_begin(f), ie = inst_end(f); i != ie; ++i) >> { >> Instruction* pDef = &(*i); >> errs() << "Def: " << *pDef << "\n"; >> >> for (User* pUser : pDef->users()) >> { >> Instruction* pUse = dyn_cast<Instruction>(pUser); >> errs() << " Use: \t" << *pUse << "\n"; >> } >> } >> >> Also, the code is executed inside a function pass. So was initially >> thinking I somehow screwed up the use information in a previous pass. >> However, I would assume the range-based iterator would not work as well >> but it does. >> >> Finally, I'm currently using LLVM 3.5.1 built for Windows. Google hasn't >> been much help. Anybody have any suggestions as to why the first example >> above doesn't work? >> >> Thanks, >> Zack >> >> >> _______________________________________________ >> LLVM Developers mailing list >> LLVMdev at cs.uiuc.edu http://llvm.cs.uiuc.edu >> http://lists.cs.uiuc.edu/mailman/listinfo/llvmdev >> >> > -- > Jon Roelofs > jonathan at codesourcery.com > CodeSourcery / Mentor Embedded >-------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.llvm.org/pipermail/llvm-dev/attachments/20150609/66e1331e/attachment.html>