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Re: [Qemu-block] [Qemu-devel] [PATCH v3 2/5] qapi: Add qobject_is_equal(

From: Eric Blake
Subject: Re: [Qemu-block] [Qemu-devel] [PATCH v3 2/5] qapi: Add qobject_is_equal()
Date: Wed, 5 Jul 2017 11:29:28 -0500
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:52.0) Gecko/20100101 Thunderbird/52.2.0

On 07/05/2017 11:22 AM, Max Reitz wrote:

>>>>     return (double)x == x && x == y;
>>> Yes, that would do, too; and spares me of having to think about how well
>>> comparing an arbitrary double to UINT64_MAX actually works. :-)
>> On second thought, this won't do, because (double)x == x is always true
>> if x is an integer (because this will implicitly cast the second x to a
>> double, too). However, (uint64_t)(double)x == x should work.
> Hm, well, the nice thing with this is that (double)UINT64_MAX is
> actually UINT64_MAX + 1, and now (uint64_t)(UINT64_MAX + 1) is
> undefined... Urgs.

(uint64_t)(UINT64_MAX + 1) is well-defined - it is 0.

(Adding in unsigned integers is always well-defined - it wraps around on
mathematical overflow modulo the integer size.  You're thinking of
overflow addition on signed integers, which is indeed undefined)

> So I guess one thing that isn't very obvious but that should *always*
> work (and is always well-defined) is this:
> For uint64_t: y < 0x1p64 && (uint64_t)y == x
> For int64_t: y >= -0x1p63 && y < 0x1p63 && (int64_t)y == x

That's harder to read, compared to the double-cast method which is
well-defined after all.

> I hope. :-/
> (But finally a chance to use binary exponents! Yay!)

Justifying the use of binary exponents is going to be harder than that,
and would need a really good comment in the code, compared to just using
a safe double-cast.

Eric Blake, Principal Software Engineer
Red Hat, Inc.           +1-919-301-3266
Virtualization:  qemu.org | libvirt.org

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