How to resolve qbyteA[5] = qbyteB?

[hiuao]

Hi,every one!Can you tell me how to resolve the following problem?

QByteArray qbyteA,qbyteB;
qbyteA.resize(9);
........
qbyteA[5] = qbyteB;//error: QByteArray:: operator QNoImplicitBoolCast() const' is private

QByteArray qbyteA,qbyteB;
qbyteA.resize(9);
........
qbyteA[5] = qbyteB;//error: QByteArray:: operator QNoImplicitBoolCast() const' is private

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I don't know why,please anyone if you know tell me,thank you very much!

[aamer4yu]

What are you trying to achieve ?
qByteA[5] = qByteB ; Here you are trying to assign the byte array qByteB to sixth character of qByteA (qByteA[5]) . How can this be possible ??
You can assign character to character. If qByteA was an array of pointers to QByteArray, then that statement would have worked.

Hence its giving error that it cannot find any suitable conversion for the statement.

[hiuao]

Thank you for your kind help,but I have already try for many ways,such as

QByteArray byteA[9],byteB; or QByteArray *byteA,byteB;
byteA[0] = 0x23;//error: invalid conversion from `int' to `const char*'

QByteArray byteA[9],byteB; or QByteArray *byteA,byteB;
byteA[0] = 0x23;//error: invalid conversion from `int' to `const char*'

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I want to give

byteA[i] = 0xxx,byteA[j] = byteB

byteA[i] = 0xxx,byteA[j] = byteB

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,how I can do? Please!

[aamer4yu]

You can assign hex values to QByteArray,,, like qByteA[5] = 0xA2 - Hex format.
or u can use QByteArray::fromHex.

If you are trying to assign some part of qByteB to qByteA then trim qByteB first and use the assignement operator.

[hiuao]

Why?But there no member QByteArray::fromHex.

[jacek]

QByteArray byteA[9],byteB; or QByteArray *byteA,byteB;
byteA[0] = 0x23;//error: invalid conversion from `int' to `const char*'

Here byteA is an array of QByteArrays, so byteA[0] is QByteArray.

I want to give byteA[i] = 0xxx,byteA[j] = byteB,how I can do?

Make sure the types match:

QByteArray byteA, byteB;
...
byteA[0] = 0x11;
byteA[1] = byteB[2];

QByteArray byteA, byteB;
...
byteA[0] = 0x11;
byteA[1] = byteB[2];

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[aamer4yu]

Why?But there no member QByteArray::fromHex.

Why ? did u click on the link ? Its a static member function.

[hiuao]

Sorry to replay so late.I think your version maybe QT4.3,but mine is QT4.2.2,so there is no fromHex in my version.I want to do as the following,please tell me if I am right,Thank you very much!

QByteArray bytA[3],bytB;
bytA[0] = "0x34";
bytA[1] = bytB;
bytA[2] = QByteArray( ( 2*(bytA[0].toLong())& 0x000000ff )<<8 );

QByteArray bytA[3],bytB;
bytA[0] = "0x34";
bytA[1] = bytB;
bytA[2] = QByteArray( ( 2*(bytA[0].toLong())& 0x000000ff )<<8 );

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I want to know if the function use "QByteArray::toLong(),& ,<<" is right.

[hiuao]

Here byteA is an array of QByteArrays, so byteA[0] is QByteArray.

Make sure the types match:

QByteArray byteA, byteB;
...
byteA[0] = 0x11;
byteA[1] = byteB[2];

QByteArray byteA, byteB;
...
byteA[0] = 0x11;
byteA[1] = byteB[2];

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I don't think its proper,excuse me to say so.If byteB only have one byte,how can it do?

[jacek]

If byteB only have one byte,how can it do?

Even if it contains a single byte, still its type is QByteArray. You have to extract that single byte in order to make the assignment:

byteA[2] = byteB[0];

byteA[2] = byteB[0];

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