Converting a QByteArray in Hexa format into an Int

[mpergand]

@Pantoufle said in Converting a QByteArray in Hexa format into an Int:

but then the QByteArray will be 8 bytes long,

Wrong, two hex digits = 1 byte.

[jsulm]

@Pantoufle Sorry, I don't understand your problem. As @mpergand wrote two hex digids represent one byte. Did you actually try toInt? And are you sure that what you receive is actually hex encoded number-string? How exactly are those numbers sent?

[JonB]

@Pantoufle
Decide whether your input is:

• 4 bytes of (binary) data, i.e. a binary representation of ant int, whether is low- or high-endian order; or
• 8 hexadecimal characters to be converted to an int (presumably high-to-low).

Then you/we will know what you're actually asking. Till then it is quite unclear, and only you know what your input is....

[Pantoufle]

@mpergand

I know that two hex digits = 1 byte, therefore in my example if i remove the fromHex it create a 8 bytes QbyteArray :

QByteArray mydata ="00004A9E";

std::cout << "My data size : " << mydata.size() << std::endl;

Output:

My data size : 8

Maybe i miss understood the "You don't need the fromHex" quote and I'm sorry if so.

@JonB

It's a 4 bytes if binary data representation of an int, high endian

[Pantoufle]

I've found an answer that works :

QByteArray mydata = QByteArray::fromHex("00004A9E");
QDataStream ds(mydata);
ds.setByteOrder(QDataStream::BigEndian);
int mydataInteger;
ds >> mydataInteger;
std::cout << "Int content : " << mydataInteger << std::endl;

But I would be glad if you can make me understand why toInt() don't work.

[jsulm]

@Pantoufle said in Converting a QByteArray in Hexa format into an Int:

why toInt() don't work

This code prints 19102:

QByteArray n("00004A9E");
bool ok;
qDebug() << n.toInt(&ok, 16);

[Pantoufle]

@jsulm

Yes but the code u just send is a 8 bytes QbyteArray :

QByteArray mydata("00004A9E");
    std::cout << "My data : " << mydata.size() << std::endl;

    bool ok;
    int a = mydata.toInt(&ok, 16);
    if(!ok)
        std::cout << "Fail convert" << std::endl;

    std::cout << "My data int " << a << std::endl;

[Pantoufle]

@Pantoufle

Forgot the output :

My data : 8
My data int : 19102

[mpergand]

@Pantoufle
What is the format of the original data ?
ASCII in hex format ?
Binary in big endian ?
Other ?

[J.Hilk]

@Pantoufle this is probably through the ambiguity of the QByteArray constructor, wich builds down to char* or an other QByteArray

use the proper constructor from a String, and it should be fine:

auto mydata = QString("00004A9E").toLatin1();
    std::cout << "My data : " << mydata.size() << std::endl;

    bool ok;
    int a = mydata.toInt(&ok, 16);
    if(!ok)
        std::cout << "Fail convert" << std::endl;

    std::cout << "My data int " << a << std::endl;

[Pantoufle]

@mpergand said in Converting a QByteArray in Hexa format into an Int:

@Pantoufle
What is the format of the original data ?
ASCII in hex format ?
Binary in big endian ?
Other ?

Binary in big endian

@J-Hilk It works but same problem as before, the QByteArray generated by the line (auto mydata = QString("00004A9E").toLatin1();) is 8 bytes long.

[Christian Ehrlicher]

@Pantoufle said in Converting a QByteArray in Hexa format into an Int:

the QByteArray generated by the line (auto mydata = QString("00004A9E").toLatin1();) is 8 bytes long.

But that's correct - you pass 8 bytes to a QString ctor and convert it back to a QByteArray.

[jsulm]

@Pantoufle said in Converting a QByteArray in Hexa format into an Int:

Binary in big endian

Then why do you pass a string to QByteArray?
You should rather use https://doc.qt.io/qt-6/qbytearray.html#fromRawData
Also see https://doc.qt.io/qt-6/qtendian.html

[Pantoufle]

@Christian-Ehrlicher

My case is equivalent to :

QByteArray mydata = QByteArray::fromHex("0000049E");

Which generate a 4 bytes array, so yeah it's normal that it generates a 8 bytes in the previous example but just it's not what I want.

@jsulm

QByteArray mydata = QByteArray::fromHex("0000049E");
Represent a binary in big endian, 4 bytes long.
I didn't know about QtEndian and fromRawData, I will check that thx.
Also I found that it works with QByteArrayLiteral("\x00\x00\x04\x9E")

[JonB]

@Pantoufle
Hi. I don't understand if you still have a problem or a question? You want a 4 byte integer, and you have one. I don't understand what the relevance is of a string you choose to pass to QByteArray::fromHex() as a convenient way of getting the 4 bytes into memory, who cares what the string is or how long it is? It has nothing to do with how your device works, which does not use hex or strings, it just sends a 4-byte integer. Which should be convertible via QByteArray::toInt() provided the byte order matches your machine's endian order (else it would need swapping).

My bad, see @Christian-Ehrlicher's clarification of QByteArray::toInt() below!

[Christian Ehrlicher]

@JonB said in Converting a QByteArray in Hexa format into an Int:

Which should be convertible via QByteArray::toInt() provided the byte order matches your machine's endian order (else it would need swapping).

No, you're wrong here.
QByteArray::toInt() converts a string representation of a number to an integer.

int32_t val = *(reinterpret_cast<int32_t*>(ba.constData()))

if the endian is the same on the source and destination, otherwise use the qEndian helper functions.

[JonB]

@Christian-Ehrlicher
Oh, I am so wrong, I should have looked at QByteArray::toInt() docs before answering instead of assuming it was a binary operation! I now understand where the confusion lies totally! I will cross out my earlier.

[Paul Colby]

Just use the Qt endian functions every time, since they evaluate to a no-op (ie static_cast) when the source and destination match anyway. Eg:

const QByteArray mydata = QByteArray::fromHex("00004A9E");
const quint32 mydataInteger = qFromBigEndian<quint32>(mydata);
qDebug() << mydata;
qDebug() << mydataInteger;

Outputs:

"\x00\x00J\x9E"
19102

Cheers.