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

[Douglas do Couto Teixeira]

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.


> 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
>>
>>
>>
>>
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>>
>
>
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