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Date: Mon, 20 Jan 2020 00:02:04 +0300
From: Alexander Cherepanov <>
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 

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 

Take floating-point zeroes for example. They are equal but have 
different provenances: one came from the right, another one -- from the 

> 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 

Alexander Cherepanov

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