[PATCH] D97915: [Coroutines] Handle overaligned frame allocation

[Chuanqi Xu via Phabricator via cfe-commits]

ChuanqiXu added a comment.

This code snippets confused me before:

  coro.alloc.align:                                 ; preds = %coro.check.align
    %mask = sub i64 %11, 1
    %intptr = ptrtoint i8* %call to i64
    %over_boundary = add i64 %intptr, %mask
    %inverted_mask = xor i64 %mask, -1
    %aligned_intptr = and i64 %over_boundary, %inverted_mask
    %diff = sub i64 %aligned_intptr, %intptr
    %aligned_result = getelementptr inbounds i8, i8* %call, i64 %diff

This code implies that `%diff > 0`. Formally, given `Align = 2^m, m > 4` and `Address=16n`, we need to prove that:

  (Address + Align -16)&(~(Align-1)) >= Address

`&(~Align-1)` would make the lowest `m` bit to 0. And `Align-16` equals to `2^m - 16`, which is `16*(2^(m-4)-1)`. Then `Address + Align -16` could be `16*(n+2^(m-4)-1)`.
Then we call `X` for the value of the lowest `m` bit of  `Address + Align -16`. 
Because `X` has `m` bit, so `X <= 2^m - 1`. Noticed that `X` should be 16 aligned, so the lowest 4 bit should be zero.
Now,

  X <= 2^m - 1 -1 - 2 - 4 - 8 = 2^m - 16

So the inequality we need prove now should be:

  16*(n+2^(m-4)-1) - X >= 16n

Given X has the largest value wouldn't affect the inequality, so:

  16*(n+2^(m-4)-1) - 2^m + 16 >= 16n

which is very easy now.

The overall prove looks non-travel to me. I spent some time to figure it out. I guess there must be some other people who can't get it immediately. I strongly recommend to add comment and corresponding prove for this code.


Repository:
  rG LLVM Github Monorepo

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