[LLVMdev] Does current LLVM target-independent code generator supports my strange chip?
Evan Cheng
evan.cheng at apple.com
Tue Nov 18 10:01:27 PST 2008
Why not model each channel as a separate physical register?
Evan
On Nov 17, 2008, at 6:36 AM, Wei wrote:
> I have a very strange and complicate H/W platform.
> It has many registers in one format.
> The register format is:
>
> ------------------------------
> ----------------------------------------------------------------------------------------
> | 24-bit | 24-bit
> | 24-bit | 24-bit |
> ----------------------------------------------------------------------------------------------------------------------
> a
> b
> c d
>
> There are 4 channels in a register, and each channel contains 24-
> bit, hence, there are total 96-bit in 'one' register.
> You can store a 24-bit integer or a s7.16 floating-point data into
> each channel.
> You can name each channel 'a', 'b', 'c', 'd'.
>
> Here is an example of the operation in this H/W platform:
>
> ADD R3.ab, R1.abab, R2.bbaa
>
> it means
>
> Add 'abab' channel of R1 and 'bbaa' channel of R2, and
> put the result into the 'ab' channel of R3.
>
> It's complicate.
> Imagine a non-existed temp register named 'Rt1', the content of its
> 'a','b','c','d' channel are got from 'a','b','a','b' channel of R1,
> and imagine another non-existed temp register named 'Rt2', the
> content of its 'a','b','c','d' channel are got from 'b','b','a','a'
> channel of R2.
> and then add Rt1 & Rt2, put the result to R3
> this means
> the 'a' channel of R3 will be equal to the 'a' channel of Rt1 plus
> the 'a' channel of Rt2, (i.e. 'a' from R1 + 'b' from R2, because
> R1.'a'bab and R2.'b'baa)
> the 'b' channel of R3 will be equal to the 'b' channel of Rt1 plus
> the 'b' channel of Rt2, (i.e. 'b' from R1 + 'b' from R2, because
> R1.a'b'ab and R2.b'b'aa)
> the 'c' channel of R3 will be untouched, the value of the 'c'
> channel of Rt1 plus the 'c' channel of Rt2 (i.e. 'a' from R1 + 'a'
> from R2, because R1.ab'a'b and R2.bb'a'a) will be lost.
> the 'd' channel of R3 will be untouched, too. The value of the 'd'
> channel of Rt1 plus the 'd' channel of Rt2 (i.e. 'b' from R1 + 'a'
> from R2, because R1.aba'b' and R2.bba'a') will be lost, too.
>
> I don't know whether I can set the 'type' of such register using a
> llvm::MVT::SimpleValueType?
> According the LLVM doc & LLVM source codes, I think llvm::MVT::v8i8,
> v2f32, etc is used to represent register for SIMD instructions.
> I don't think the operations in my platform are SIMD instructions.
> However, I can not find any llvm::MVT::SimpleValueType which can
> represents a 96-bit register.
>
> Thus, my question is:
>
> 1) Does current LLVM backend supports this H/W?
> 2) If yes, how can I write the type of the register class in my .td
> file?
>
> (Which value should I fill in the following 'XXX' ?)
> def TempRegs : RegisterClass<"MFLEXG", [XXX], 32, [R0, R1, R2, R3,
> R4, R5, R6, R7, R8, R9,
> R10, R11, R12,
> R13, R14, R15, R16, R17, R18, R19,
> R20, R21, R22,
> R23, R24, R25, R26, R27, R28, R29,
> R30, R31]> {
> }
>
> 3) If not, does this means I have to write the whole LLVM backend
> based on the basic llvm::TargetMachine & llvm::TargetData, just like
> what CBackend does?
>
>
> --------------------------------------------------------
> Wei Hu
> http://www.csie.ntu.edu.tw/~r88052/
> http://wei-hu-tw.blogspot.com/
>
> _______________________________________________
> 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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