Concatenation of 2 byte arrays in constructor gives strange result-CodePudding

Qt 6.2.1 MinGW

I have 2 arrays, header FirstArray and body SecondArray. I know that copy-paste, isn't good in programming, so at first, did so :

int main(int argc, char *argv[]) 
{
    QCoreApplication a(argc, argv);
    const QByteArray FirstArray=QByteArray(std::begin<char>({static_cast<char>(0xA5), 0x5A, static_cast<char>(0xB5), 0x5B}), 4);
    const QByteArray SecondArray=QByteArray(FirstArray std::begin<char>({0x18, 0x00, 0x01, static_cast<char>(0x9B), 0x03, 0x09, 0x00, 0x19, static_cast<char>(0x91)}), 13);
    qDebug()<< SecondArray.toHex();

    return a.exec(); 
}

I expected the result:

"a55ab55b1800019b0309001991"

But in output I see:

"a55ab55b1800adbaababababab"

Then I rewrite the second QByteArray initialization, and remove plus operation in constructor:

const QByteArray SecondArray=QByteArray(std::begin<char>({static_cast<char>(0xA5), 0x5A, static_cast<char>(0xB5), 0x5B, 0x18, 0x00, 0x01, static_cast<char>(0x9B), 0x03, 0x09, 0x00, 0x19, static_cast<char>(0x91)}), 13);

I get:

"a55ab55b1800019b0309001991"

Why does it's happed in the first case? How to write correctly?

CodePudding user response：

As mentioned in a comment, you are calling

const QByteArray operator (const QByteArray &a1, const char *a2)

Which, per its documentation:

Returns a byte array that is the result of concatenating byte array a1 and string a2.

It excepts a2 to point to a null-terminated string.

You can rearrange the concatenation to

FirstArray   QByteArray(std::begin<char>({0x18, ......

To call the other overload which does not rely on null-termination of a c-string.