QByteArray copy [Solved]

sos1

I'm trying to copy some bytes from one QByteArray to another and I'm having trouble getting it to work. This should be simple but somehow...

The compile error is "error: invalid conversion from 'char' to 'const char*' "

@
void Probe::OP_DataReceive(const QByteArray qa)
{
unsigned char i;

char * ptr;
   
ptr = (char *)qa.data() + 3;
for (i=0; i<15; i++)
{
    MyData.Probe.Name[i] = *ptr++;
}

@

The destination QByteArray is set up as follows.
@
struct MyData_ {
struct Probe_{
unsigned char Number;
QByteArray Name[15];
} Probe;
unsigned char ProbeIsPresent;
} extern MyData;
@

Thanks,
James

SGaist

Hi,

I'm not sure you really understood how QByteArray works.

If you want to copy a part of a QByteArray simply use:

@QByteArray part = originalArray.mid(startingPoing, length)@

In you MyData structure you are declaring an array of QByteArray, are you sure that's what you want ?

sos1

You are right, I thought I understood but now I'm not sure.

Can you tell me why this doesn't work?
@
for (i=0; i<15; i++)
{
MyData.Probe.Name[i] = *ptr++;
}
@

No, you are correct, I did not want an array of "Name". So, in the structure how should I reserve 15 bytes for the name?

Thanks,
James

SGaist

MyData.Probe.Name[i] returns the QByteArray at i,

*ptr++ returns the char pointed by ptr and then increments ptr.

Jeroentjehome

What is wrong with the = operator for QByteArray?
and as SGalst said, use other special options to only copy parts of the original QByteArray.
BTW the QByteArray is not like a C array. The class itself will allocate or release required memory, so no need to add the size in the struct. Using the [] operators should be done with care! It is easy to have buffer overflows with them!

MuldeR

@QByteArray Name[15]@

...creates 15 separate QByteArrays. It's an array of QByteArray's.

--

Instead try:

@void foo(const QByteArray &qa)
{
const char* ptr = qa.constData() + 3;

QByteArray myByteArray(15);
char *outPos = myByteArray.data();

for (i=0; i<15; i++)
{
*outPos++ = *ptr++;
}
}@

--

But I think you can also do:

@void foo(const QByteArray &qa)
{
QByteArray myByteArray;
myByteArray = qa.mid(3, 15);
}@

SGaist

MuldeR, based on your code and if Name becomes a simple QByteArray.

wouldn't

@
MyData.Probe.Name = qa.mid(3, 15);
@

be cleaner ?

sos1

Thanks very much all!

James

Jeroentjehome

Plz add [SOLVED] to the first post!