Date: Mon, 20 Jan 2020 00:02:04 +0300 From: Alexander Cherepanov <ch3root@...nwall.com> To: musl@...ts.openwall.com Subject: Re: Minor style patch to exit.c On 19/01/2020 19.22, Rich Felker wrote: >> Even if we know that _start + k == _end it doesn't mean that we >> allowed to subtract them. > > Consider a function that takes a pointer p, an array a, and a length > l, and does: > > for (i=0; i<l; i++) if (a+i == p) return p-a; > > Can f(_end,_start,k) and f(_start+k,_start,k) legitimately differ, > despite _end==_start+k? I guess it depends on what you mean by "legitimately" and "differ". Given that _start and _end are different arrays the first variant is undefined. Counter-intuitive behavior of equal pointers could be demonstrated much easier. Suppose x and y are two objects of the same type and &x + 1 == &y. Is it valid to evaluate the following expressions: *(&x + 1), &x + 2, (&y)[-1]? > I think the answer is no, in the existing C > language, in that the result of an expression is a pure function of > the *values* put into it. The fact that two values are equal doesn't mean that they are the same value. Take floating-point zeroes for example. They are equal but have different provenances: one came from the right, another one -- from the left:-) > But compiler folks do not want to interpret > it this way and are pushing through hidden "provenance" state, so... IIUC they are not happy about it too but the alternatives are not that great. -- Alexander Cherepanov
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