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Date: Sun, 19 Jan 2020 14:49:56 +0000
From: Pascal Cuoq <>
To: "" <>
Subject: Re: Minor style patch to exit.c

> On 19 Jan 2020, at 15:24, Markus Wichmann <> wrote:
> That reminds me of something I read in the C standard: Two pointers must
> compare equal if, among other possibilities, one is a pointer to
> one-past its underlying array, and the other is a pointer to the start
> of its array, and the arrays happen to lie behind one another in address
> space.

The clause is 6.5.9:6 in C11:

Two pointers compare equal if and only if both are null pointers, both are pointers to the same object (including a pointer to an object and a subobject at its beginning) or function, both are pointers to one past the last element of the same array object, or one is a pointer to one past the end of one array object and the other is a pointer to the start of a different array object that happens to immediately follow the first array object in the address space.

With a footnote:

Two objects may be adjacent in memory because they are adjacent elements of a larger array or adjacent members of a structure with no padding between them, or because the implementation chose to place them so, even though they are unrelated. […]

The way GCC developers have decided to interpret this clause is that if the two arrays had to be adjacent in memory because they were part of the same aggregate, the pointers will reliably be equal iff one is one-past-the-end of the first one and the other is at the beginning of the second one. Otherwise, compilers will sometimes optimize “&a + 1 == b” to 0, that is, they implement a rule that could be paraphrased as “two pointers *may* compare equal if one is to the end of an array object and the other is to the beginning of another array objects that happens to lie immediately after the first one”.

> Therefore, if _start and _end were arrays, even the GCC devs must agree
> that there might be an integer i such that _start + i == _end.

If they did, they would not have written an optimization that transforms “a + 1 == b” into 0 regardless of the actual addresses of a and b.

(Every object is an array for the purpose of this discussion, that's 6.5.9:7, but it doesn't help.)


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