QString::toShort problem

JonB

@Bart_Vandewoestyne
There appears to be a difference in behaviour between toShort() and toUShort() which the documentation does not mention and @kshegunov is fiendlishly taking advantage of with internal knowledge :)

Christian Ehrlicher

FFFE is simply to big for a signed short ...

Bart_Vandewoestyne

@Christian-Ehrlicher said in QString::toShort problem:

FFFE is simply to big for a signed short ...

Why? Considering Visual Studio 2015 (and I assume also a lot of other compilers), the range for (signed) short is –32,768 to 32,767 (see https://msdn.microsoft.com/nl-be/library/s3f49ktz.aspx). The value -2 (represented by FFFE = two's complement) fals nicely into that range. So that's why I was expecting to be able to go from "FFFE" to -2 using QString::toShort()...

JonB

@Bart_Vandewoestyne
Because 0xFFFE is 65,534. It fits in an unsigned short range, but overflows the signed short's maximum positive value of 32,767. That's what @Christian-Ehrlicher is saying.

You are assuming that QString::toShort() will treat the string 0xFFFE as meaning exactly the same thing as -2, but it doesn't. It regards it as a positive number which is beyond the range of signed shorts, not as an alternative way of writing -2.

jsulm

@JonB I disagree.
Complement on two for 2:

  0000 0010
  1111 1101
+ 0000 0001
  1111 1110

So, -2 is 1111 1110 or 0xFE - why should this not feet into a signed short?
"You are assuming that QString::toShort() will treat the string 0xFFFE as meaning exactly the same thing as -2, but it doesn't" - why should toShort() not treat 0xFE as -2?

kshegunov

@jsulm said in QString::toShort problem:

@JonB I disagree.

You shouldn't. ;)

why should toShort() not treat 0xFE as -2?

Quite simply because you don't have a fixed-size data field to work with as input. Why should toShort assume that you meant exactly the binary representation. You could've just as well had a data that's too big to fit the type. Say I'm reading some input and I'm trying to get it into a short. Suddenly due to an error or by whatever chance I get a number that's too big for my short, but instead of overflowing the toShort would give me an invalid value. It doesn't make sense that the person who implemented toShort would just jump the gun on such an assumption.
And lastly, what should we do with overflows of this kind - 0x100FF, shall toShort return 255 in this case?

jsulm

@kshegunov said in QString::toShort problem:

Why should toShort assume that you meant exactly the binary representation

Maybe I'm still sleeping and oversee something. What else should it assume? If I say its hex and pass FFFE - how does toShort() interpret it?
0x100FF is too big for a short and toShort() should return 0/false (and it does). But FFFE is a valid signed short number.

Bart_Vandewoestyne

@jsulm said in QString::toShort problem:

@kshegunov said in QString::toShort problem:

Why should toShort assume that you meant exactly the binary representation

Maybe I'm still sleeping and oversee something. What else should it assume? If I say its hex and pass FFFE - how does toShort() interpret it?
0x100FF is too big for a short and toShort() should return 0/false (and it does). But FFFE is a valid signed short number.

@jsulm I completely agree! (although I have the same feeling about sleeping and maybe overseeing something ;-) Maybe it's time to dive into the Qt 4.8.7 source and investigate why QString::toShort() is failing on "FFFE"? (does Qt 5.X also fail on that btw?)

kshegunov

@jsulm said in QString::toShort problem:

But FFFE is a valid signed short number.

No it isn't, and that's the point. Start doing the math in your head and see for yourself:

E * 1 + F * 16 + F * 16^2 + F * 16^3

And the last term overflows, which overflow is caught and voila!
If you have

char z = 127;

then:

z += 1;

Is overflowing, no matter whether the value you get is "correct".

jsulm

@kshegunov I still don't get it.
What is the representation of -2 as signed short? Isn't it 0xFFFE?

  0000 0000 0000 0010 - 2
  1111 1111 1111 1101 - invert
+ 0000 0000 0000 0001 - add 1
  1111 1111 1111 1110
-> 0xFFFE

JKSH

OK, here's an exercise to settle the debate. First, assume that QString::toShort() behaves exactly as you expect.

What should each QString (p_*) be initialized to, in order to get 32 for every output line?

QString p_oct, p_dec, p_hex, p_r32;

// ... Initialize QStrings here ...

qDebug() << p_dec.toShort(nullptr, 10); // Returns 32
qDebug() << p_hex.toShort(nullptr, 16); // Returns 32

qDebug() << p_oct.toShort(nullptr, 8);  // Returns 32
qDebug() << p_r32.toShort(nullptr, 32); // Returns 32

Next, what should each QString (n_*) be initialized to, in order to get -32 for every output line?

QString n_oct, n_dec, n_hex, n_r32;

// ... Initialize QStrings here ...

qDebug() << n_oct.toShort(nullptr, 8);  // Returns -32
qDebug() << n_dec.toShort(nullptr, 10); // Returns -32
qDebug() << n_hex.toShort(nullptr, 16); // Returns -32
qDebug() << n_r32.toShort(nullptr, 32); // Returns -32

Decide on your answer for all 8 strings first, then post your answer here.

JonB

@jsulm , @Bart_Vandewoestyne

I don't get what you don't get about: 0xFFFE is a positive overflow for parsing & storing into a ushort. Hence the behaviour.

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.

Nobody has looked at it "the other way round". I cannot test because I am Python/PyQt not C++, but what does

QString("-2").toUShort(&ok, 16)

return? In your theory it should be 0xFFFE, but I am "hoping"(!) it returns an error, just like QString("FFFE").toShort(&ok, 16) does?

Assuming that is the case, this means we do not have an ambiguity/duplication, whereby both FFFE and -2 strings can be parsed as the same number by toShort()/toUShort() (but 2 is the only way to write +2).

J.Hilk

toShort makes a toLongLong interpretation first and than casts it to short theres where the "error" comes from:

short QString::toShort(bool *ok, int base) const
{
    long v = toLongLong(ok, base);
    if (v < SHRT_MIN || v > SHRT_MAX) {
        if (ok)
            *ok = false;
        v = 0;
    }
    return (short)v;
}

toLongLong will return ‭65534‬, (0xFFFE in int64 is positve after all), and that is bigger than SHRT_MAX -> 0 and failed conversion

jsulm

@JonB said in QString::toShort problem:

In your theory it should be 0xFFFE

No, it would not, because -2 is not a hex number...
"I don't get what you don't get about: 0xFFFE is a positive overflow for parsing & storing into a ushort" - we are not talking about unsigned short, but signed short and 0xFFFE is the representation of -2.

JonB

@J.Hilk
In that case, try passing something like 0xFFFFFFFE or 0xFFFFFFFFFFFFFFFE for the string to toShort() and those who want -2 instead of error should get it?!

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