[LLVMdev] How to change the type of an Instruction?

[Nick Lewycky]

On 24 January 2011 10:05, Douglas do Couto Teixeira <
douglasdocouto at gmail.com> wrote:

>
>
> On Mon, Jan 24, 2011 at 3:01 PM, Nick Lewycky <nicholas at mxc.ca> wrote:
>
>> On 01/24/2011 04:41 AM, Douglas do Couto Teixeira wrote:
>>
>>> Hi,
>>>
>>> Nick, thanks for the reply.
>>> I still have a problem: I only need to "clone" an Instruction, changing
>>> its type. That is, I would like to keep all characteristics of the old
>>> Instruction and create a new one only with a different type.
>>>
>>
>> Sure, but what about its operands? An "add" instruction must have the same
>> type as its operands, what do you want to do with them?
>
>
> I also need to convert the type of the operands. But I want to do this when
> they are created instead of inserting "trunc" instructions before performing
> an operation. But it seems hard to me.
>

It is. You can walk upwards in a chain:
 a) if it's a constant, use ConstantExpr::getTrunc which will return another
constant.
 b) if it's a memory load you can try to replace the pointer type too and
recurse on that. But then what do you do if it's a load of a global
variable?
 c) what if it's the type returned by a function call? Do you have a
different function you can call instead? Are you going to recursively
transform that function too?
 d) what if it's an Argument? You'll have to change the function's type
which means creating a new function and splicing all the basic blocks from
the old one into a new one -- then updating every call site.
And don't forget that for each instruction you touch, you need to update all
of their users too.

Nick

Suppose you're going from a 32-bit add to a 64-bit add, do the old operands
>> get zero extended? Sign extended? You'll need to insert instructions for
>> that (unless they're constants in which case you can use constant
>> expressions). Similarly, what if the old type is a float and the new one is
>> an int? float to signed int, float to unsigned int, or bitcast (only legal
>> sometimes)?
>
>
> I believe I don't need worry about it because I only create smaller
> instructions. So I never convert a 32-bit instruction in a 64-bit
> instruction.
>
>
>>
>>
>>  I am trying
>>
>>> create a new Instruction thus:
>>>
>>> %3 = add nsw i32 %1, %2 ; <i16> [#uses=2]  //Old Instruction
>>>
>>> Value* Op0 = I->getOperand(0);
>>> Value* Op1 = I->getOperand(1);
>>> Value* V0 = new Value(Type::getInt16Ty(Op0->getContext()),
>>> Op0->getValueID());
>>>
>>
>> Hunh, Value's constructor is protected.
>>
>> In any event, Value is pure base. Constructing one this way will never get
>> you what you want. If the ValueID indicates an Instruction, go through
>> Instruction to create one.
>>
>>
>>  Value* V1 = new Value(Type::getInt16Ty(Op1->getContext()),
>>> Op1->getValueID());
>>> Instruction* newInst = BinaryOperator::CreateNSWAdd(V0, V1, "test");
>>> errs() << "NewInst:\n" << *newInst << "\n";
>>>
>>>
>>> But I get something like this:
>>>
>>> %test = add nsw i16 <badref>, <badref> ; <i16> [#uses=0]
>>>
>>
>> The two instructions V0 and V1 you created were never inserted into the
>> BasicBlock so they can't be numbered, and also they don't have names.
>>
>>
>>  What I am doing wrong?
>>>
>>
>> Suppose that you're going from i32 to i16. Your only choice with that
>> particular pair of types is a truncate. So:
>>
>>  IRBuilder builder(OldInst);
>>  Value *V0 = builder.CreateTrunc(Op0, Type::getInt16Ty());
>>  Value *V1 = builder.CreateTrunc(Op1, Type::getInt16Ty());
>>  Value *Add = builder.CreateNSWAdd(V0, V1, "test");
>>
>> The IRBuilder will take care of the distinction between instructions and
>> constants for you. Note that I have not tested the above code, it may need
>> some fixing before it compiles.
>>
>> Nick
>>
>>
>>>
>>> Best,
>>>
>>> Douglas
>>>
>>> On Fri, Jan 21, 2011 at 8:25 PM, Nick Lewycky <nlewycky at google.com
>>> <mailto:nlewycky at google.com>> wrote:
>>>
>>>    On 21 January 2011 12:56, Douglas do Couto Teixeira
>>>    <douglasdocouto at gmail.com <mailto:douglasdocouto at gmail.com>> wrote:
>>>
>>>        Hello guys,
>>>
>>>        I wonder how I can change the type of an integer variable. For
>>>        instance, given the instruction "%3 = add i32 %1, %2" I would
>>>        like to alter the instruction to "%3 = add i16 %1, %2". Is there
>>>        any way to do this?
>>>
>>>
>>>    No. Instead you create a new Instruction, in this case with
>>>    BinaryOperator::CreateAdd, then OldInst->replaceAllUsesWith(NewInst)
>>>    to update all the users, then OldInst->eraseFromParent() since it's
>>>    now dead code.
>>>
>>>    Also, all values have types immutably assigned at creation, so
>>>    you'll need to insert casts (trunc instructions in your case) to
>>>    cast %1 and %2 from i32 to i16 for the smaller add.
>>>
>>>    Nick
>>>
>>>
>>>
>>>
>>> _______________________________________________
>>> LLVM Developers mailing list
>>> LLVMdev at cs.uiuc.edu         http://llvm.cs.uiuc.edu
>>> http://lists.cs.uiuc.edu/mailman/listinfo/llvmdev
>>>
>>
>>
>
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