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> It can't if arbitrary code (for example inside bar()) can forge pointers to that memory location. Yes. It absolutely can. What are you even talking about? C
by torstenvl 1y ago
> It can't if arbitrary code (for example inside bar()) can forge pointers to that memory location.
Yes. It absolutely can. What are you even talking about?
C is not the Windows Start Menu. This habit of thinking it needs to do what it thinks I might expect instead of what I told it is deeply psychotic.
- gpderetta 1y agoI litterally have no idea what are you trying to say. Do you mean that bar should be allowed to access *ptr with impunity or not?
- torstenvl 1y agoI'm not trying to say anything. I said and meant exactly what I said. No more, no less. Your logic is obviously flawed. There is nothing preventing that optimization in the presence of a forged pointer in bar().
- gpderetta 1y agoEither there is no provenance, forging is allowed and the optimization is disallowed; or there is provenance and forging the pointer and attempting to inspect (or modify) the value of *ptr in bar() is UB.
- torstenvl 1y agoAttempting to inspect or modify the value of *ptr in bar() through a forged pointer was always UB. You are saying absolutely nothing meaningful.
- ncruces 1y agoYou never converted ptr to an integer. If you did, if the pointer escapes, yes, I claim that then the allocation can't be optimized away. Why is that so bad?