From: Ralf Jung Date: Mon, 14 Dec 2020 17:53:27 +0000 (+0100) Subject: fix int->uintptr_t fallout X-Git-Url: https://git.ralfj.de/web.git/commitdiff_plain/e085d6a060034703d21ebe36d9bf964eb4af1af8?hp=e94d36e02fceb340bbd194d908b5bef527663acb fix int->uintptr_t fallout --- diff --git a/personal/_posts/2020-12-14-provenance.md b/personal/_posts/2020-12-14-provenance.md index 9d4ef2c..695e00a 100644 --- a/personal/_posts/2020-12-14-provenance.md +++ b/personal/_posts/2020-12-14-provenance.md @@ -187,7 +187,7 @@ The second optimization gives us a clue into this aspect of LLVM IR semantics: c To see why, consider the two expressions `(char*)(uintptr_t)(p+1)` and `(char*)(uintptr_t)q`: if the optimization of removing pointer-integer-pointer roundtrips is correct, the first operation will output `p+1` and the second will output `q`, which we just established are two different pointers (they differ in their provenance). The only way to explain this is to say that the input to the `(char*)` cast is different, since the program state is otherwise identical in both cases. -But we know that the integer values computed by `(uintptr_t)(p+1)` and `(uintptr_t)q` (i.e., the bit pattern of length 32 that serve as input to the `(char*)` casts) are the same, and hence a difference can only arise if these integers consist of more than just this bit pattern---just like pointers, integers have provenance. +But we know that the integer values computed by `(uintptr_t)(p+1)` and `(uintptr_t)q` (i.e., the bit pattern as represented in some CPU register) are the same, and hence a difference can only arise if these integers consist of more than just this bit pattern---just like pointers, integers have provenance. Finally, let us consider the first optimization. Here, a successful equality test `iq == ip` prompts the optimizer to replace one value by the other.