llvm dev - Sep 2017 - What should a truncating store do?

For example, truncating store of an i32 to i6. My assumption was that this
should write the low six bits of the i32 to somewhere in memory.

Should the top 24 bits of a corresponding 32 bit region of memory be
unchanged, zero,  undefined?

Should the two bits that would round the i6 up to a byte be preserved,
zero, undefined?

I can't write six bits directly so am trying to determine what set of
bitwise ops to apply between a load and subsequent store to emulate the
truncating store.

Thanks!

Jon
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On 9/15/2017 5:49 AM, Jon Chesterfield via llvm-dev
wrote:
> For example, truncating store of an i32 to i6. My assumption was that 
> this should write the low six bits of the i32 to somewhere in memory.
>
> Should the top 24 bits of a corresponding 32 bit region of memory be 
> unchanged, zero,  undefined?
Unchanged.
> Should the two bits that would round the i6 up to a byte be preserved, 
> zero, undefined?
Zero.  Legalization will normally handle this for you, though, by 
transforming it to an i8 store.

-Eli

-- 
Employee of Qualcomm Innovation Center, Inc.
Qualcomm Innovation Center, Inc. is a member of Code Aurora Forum, a Linux
Foundation Collaborative Project

On 09/15/2017 10:55 AM, Friedman, Eli via llvm-dev
wrote:
> On 9/15/2017 5:49 AM, Jon Chesterfield via llvm-dev wrote:
>> For example, truncating store of an i32 to i6. My assumption was that
this should write the low six bits of the i32 to somewhere in memory.
>>
>> Should the top 24 bits of a corresponding 32 bit region of memory be
unchanged, zero,  undefined?
> 
> Unchanged.
> 
>> Should the two bits that would round the i6 up to a byte be preserved,
zero, undefined?
> 
> Zero.  Legalization will normally handle this for you, though, by
transforming it to an i8 store.
> 
Why is this Zero?  The language ref says the value of those bits are
unspecified.

-Tom
> -Eli
>