[PATCH] D53949: [clang][CodeGen] Implicit Conversion Sanitizer: discover the world of CompoundAssign operators

[Roman Lebedev via Phabricator via cfe-commits]

lebedev.ri added a comment.

In https://reviews.llvm.org/D53949#1282884, @regehr wrote:

> I do not agree that ++ is performed on the original type. The C99 standard (6.5.3.1.2) appears to be very clear on this point: "The expression ++E is equivalent to (E+=1)."




In https://reviews.llvm.org/D53949#1282911, @rsmith wrote:

> > In https://reviews.llvm.org/D53949#1282884, @regehr wrote:
> > 
> >> The C99 standard (6.5.3.1.2) appears to be very clear on this point: "The expression ++E is equivalent to (E+=1)."
> > 
> > 
> > That is clearly not what clang is doing here.
>
> How so? E += 1 is equivalent to E = E + 1, which effectively performs the arithmetic in the promoted type, which is what clang (effectively) does. (The potential promotion from E to int can't change the value, but the implicit conversion from the type of E + 1 to the type of E can, if E's type is not a promoted type.)


*effectively*. In other words, which sanitizer should be catching this?
https://godbolt.org/z/AFpRWS
Unless i'm misreading @regehr the promotions *should* happen, and this is supposed to be caught by *this* conversion sanitizer, not the usual overflow sanitizers.


Repository:
  rC Clang

https://reviews.llvm.org/D53949