QString::toShort problem

[JonB]

@jsulm

No, it would not, because -2 is not a hex number...

Yes it is! It's as much a hex number as some other base.

[jsulm]

@JonB said in QString::toShort problem:

In that case, try passing something like 0xFFFFFFFE or 0xFFFFFFFFFFFFFFFE for the string to toShort()

Come on - these numbers are NOT short. We should stay on topic.

[JonB]

@jsulm said in QString::toShort problem:

Come on - these numbers are NOT short. We should stay on topic.

I beg your pardon!? I am totally on topic. I was replying to @J-Hilk 's display of the code of QString::toShort(). Did you try what I suggested rather than dismissing it as OT? In view of the code shown, I am trying to suggest what 0xFFF.... string toShort() will accept as representing a negative number....

[JKSH]

Nobody wants to try my exercises... (sad face)

[jsulm]

@JonB Passing 0xFFFFFFFE returns 0

[J.Hilk]

@JonB
actually, no take a look at toLongLong

qint64 QString::toLongLong(bool *ok, int base) const
{
#if defined(QT_CHECK_RANGE)
    if (base != 0 && (base < 2 || base > 36)) {
        qWarning("QString::toLongLong: Invalid base (%d)", base);
        base = 10;
    }
#endif

    bool my_ok;
    QLocale def_locale;
    qint64 result = def_locale.d()->stringToLongLong(*this, base, &my_ok, QLocalePrivate::FailOnGroupSeparators);
    if (my_ok) {
        if (ok != 0)
            *ok = true;
        return result;
    }

    QLocale c_locale(QLocale::C);
    return c_locale.d()->stringToLongLong(*this, base, ok, QLocalePrivate::FailOnGroupSeparators);
}

I think, haven't looked stringToLongLong up, that here happens stirng lentgh magic, because every combinaion of FFF..E up to to 0xFFFFFFFE is interpretated as the uint value and everything above as -2 (as returning int64 value)

[JonB]

@jsulm

@JonB Passing 0xFFFFFFFE returns 0

Since QString::toLongLong() returns a qint64 (8 bytes, not 4), did you try 0xFFFFFFFFFFFFFFFE ?

[jsulm]

@JonB said in QString::toShort problem:

Since QString::toLongLong() returns a qint64 (8 bytes), did you try 0xFFFFFFFFFFFFFFFE ?

Returns 0 as well.
And I don't see why it should depend on the length.

[J.Hilk]

@JonB surprisingly enough

qDebug() << std::numeric_limits<int64_t>::min() << std::numeric_limits<int64_t>::max()
             << endl << (int64_t)0xFFFFFFFFFFFFFFFE;
 QString s("0xFFFFFFFFFFFFFFFE"); bool ok;
short sh =  s.toShort(&ok, 16);
qDebug() <<sh << ok;
long lg = s.toLongLong(&ok,16);
qDebug() << lg << ok;

returns:

-9223372036854775808 9223372036854775807 
-2
0 false
0 false

[JonB]

@jsulm
It would "depend on the length", as you put it, because as a 64-bit number 0xFFFFFFFE != 0xFFFFFFFFFFFFFFFE.

[jsulm]

@JonB I want to convert a signed short number not long or long long or ...
0xFFFE as signed short is -2 - do you agree (I mean independently from what Qt toShort() thinks it is)?

[jsulm]

@JonB

qDebug() << (short)0xFFFE;

prints -2 as expected

[JonB]

@jsulm
I believe the problem here is a confusion between "bit representation" and "string representation".

• It is undoubtedly, unambiguously true that, for signed short, 0xFFFE as a bit pattern is -2.
• However, for signed short, 0xFFFE as a string "could" be either -2 (which fits in a short) or 65,534 (which does not fit in a short). And QString::toShort() is taking the latter interpretation, and hence erroring.

[J.Hilk]

@jsulm said in QString::toShort problem:

@JonB

qDebug() << (short)0xFFFE;

prints -2 as expected

qDebug() << (short)0xFFFFFFFFFFFFFFFE;

prints also -2, would one expect that

actually yes, the first bytes are simply dropped x)

[JonB]

@jsulm said in QString::toShort problem:

@JonB

qDebug() << (short)0xFFFE;

prints -2 as expected

Yes, that's why I wrote earlier:

One thing that is clear: the implementation of QString::toShort() is not static_cast<short>(QString::toUShort()), even if that might have been the way you were tempted to do it.

[jsulm]

@JonB said in QString::toShort problem:

However, for signed short, 0xFFFE as a string "could" be either -2 (which fits in a short) or 65,534

No, signed short 0xFFFE is -2 even as string, because I'm calling toShort() not toUShort().
And why does

qDebug() << (short)0xFFFE;

print -2?

[jsulm]

@JonB said in QString::toShort problem:

the implementation of QString::toShort() is not static_cast<short>(QString::toUShort())

I never said that

[JonB]

@jsulm
But you're asking why qDebug() << (short)0xFFFE; prints -2. And I'm saying that's because of the way "cast-to-short" works in C++, which is simply not what the implementation of Qt's QString::toShort() does or purports to do.

Basically, "cast-to-short" ((short)) has no concept ever of "overflow/error", but QString::toShort() does have a concept of "overflow/error", and that's why they work differently. They are not intended to be equivalent.

[I am beginning to feel the need for @kshegunov 's moral support here, because I feel I am being attacked ( :( ) and it is indeed all to do with the overflowing he mentioned in his earlier reply.]

[jsulm]

@JonB What overflow error do you mean? 0xFFFE is a valid short number in both cases: signed and unsigned.

[kshegunov]

Hey, let's do it the russian way and settle this outside, huh? Take a breath people.

@jsulm
Johann, you're wrong simply because "0xFFFE" is not a negative number, but a string, that simple. I know that in 2's complement for short this is -2, but that's if you go to the actual implementation of the negative numbers. The fact of the matter is there have been implementations that do not use integer complements. This string is not a binary representation, that is all, so don't expect the function to assume it should convert in binary-like way! Otherwise, as Jonas pointed out earlier "0xFFFFFFFFFFE" should just expand to -2 as well due to truncations.