On 07/05/2017 08:48 AM, Max Reitz wrote:
>>> /**
>>> + * qnum_is_equal(): Test whether the two QNums are equal
>>> + */
>>> +bool qnum_is_equal(const QObject *x, const QObject *y)
>>> +{
>>> + QNum *num_x = qobject_to_qnum(x);
>>> + QNum *num_y = qobject_to_qnum(y);
>>> +
>>> + switch (num_x->kind) {
>>> + case QNUM_I64:
>>> + switch (num_y->kind) {
>>> + case QNUM_I64:
>>> + /* Comparison in native int64_t type */
>>> + return num_x->u.i64 == num_y->u.i64;
>>> + case QNUM_U64:
>>> + /* Implicit conversion of x to uin64_t, so we have to
>>> + * check its sign before */
>>> + return num_x->u.i64 >= 0 && num_x->u.i64 == num_y->u.u64;
>>> + case QNUM_DOUBLE:
>>> + /* Implicit conversion of x to double; no overflow
>>> + * possible */
>>> + return num_x->u.i64 == num_y->u.dbl;
>>
>> Overflow is impossible, but loss of precision is possible:
>>
>> (double)9007199254740993ull == 9007199254740992.0
>>
>> yields true. Is this what we want?
>
> I'd argue that yes, because the floating point value represents
> basically all of the values which are "equal" to it.But the problem is that we CAN represent the fully-precise number as an integer, so having multiple distinct integers that compare differently against each other, but equal to the same double, is awkward. > > But I don't have a string opinion. I guess the alternative would be to > convert the double to an integer instead and check for overflows before? That's the solution Markus gave, and I'm in favor of the tighter check: > >> I guess the obvious fix is >> >> 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. :-) It basically says that we are unwilling to declare an integer equivalent to the double if the double loses precision when trying to store the integer. -- Eric Blake, Principal Software Engineer Red Hat, Inc. +1-919-301-3266 Virtualization: qemu.org | libvirt.org
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