[llvm-dev] How exactly is datatype alignment determined?

[陳韋任 via llvm-dev]

Would http://llvm.org/docs/LangRef.html#langref-datalayout be help?

- chenwj

2017-05-22 23:26 GMT+08:00 Dr. ERDI Gergo via llvm-dev <
llvm-dev at lists.llvm.org>:

> On Mon, 22 May 2017, Dr. ERDI Gergo wrote:
>
> Actually, tracking down the sequence of function calls, it turns out that
>> '8' is ultimately coming from the following call in
>> DataLayout::getAlignment:
>>
>> getAlignmentInfo(AGGREGATE_ALIGN, 0, abi_or_pref, Ty);
>>
>> this seems to return 8 with the following datalayout string:
>>
>> e-S8:p:16:8-i8:8-i16:8-i32:8-i64:8-f32:8-f64:8-n8-a:8
>>
>> I think my problem is that 'a:8' probably doesn't mean what I think it
>> should mean. What is the difference between 'a:8' and 'a:0'?
>>
>
> OK I just now tried with 'a:0' for identical result. I am now quite sure I
> am fully confused about what these alignment settings mean.
>
> Does 'a:8' mean single byte alignment (i.e. *no* alignment)? If yes, does
> that mean getAlignment for a struct type should return 1? In fact, when
> getAlignment returns 8, does that mean 8 bits (no alignment) or 8 bytes (64
> bits) alignment? Or does that mean the lowest 8 bits of the address needs
> to be 0, i.e. 256-byte alignment?
>
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-- 
Wei-Ren Chen (陳韋任)
Homepage: https://people.cs.nctu.edu.tw/~chenwj
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