QByteArray increment, and qDebug() benaviour

McLion

Hi

A very basic one .. I just can't find it ;-)

Is it just me that is missing something or is it not possible to increment in a QByteArray?

qbTest[iIdx]+=1;

If possible, how?
Or do I better use a standard C++ array?

Something else I just can't find why:
If qbTest[iIdx] is = 1, qDebug() does output nothing - which is OK I assume because 0x01 is 'not visible'.
If I add '0' it should be 0x31 which is '1', but it outputs 49 - I regularly use this in C without issues.
If I do +'0'-48 - which actually seems very stupid, does it - it shows a correct '1'
Can someone shed some light on this?

Thanks

raven-worx

@McLion said in QByteArray increment:

Is it just me that is missing something or is it not possible to increment in a QByteArray?

i think you are just assigning your value to a copy and thus it gets lost.

Does the following work?

qbTest[iIdx] = char(qbTest[iIdx]) + 1;

McLion

@raven-worx

qbTest[iIdx] = qbTest[iIdx] + 1;

works.

McLion

@raven-worx
+= as well as ++ returns a 'no match for operator...'

raven-worx

@McLion
QByteArray's [] operator returns a QByteRef (which is unfortunately not documented, but available via public API)
Honestly i am not quite sure why the + even works, since the QByteRef class hasn't defined it. But i guess the compiler implicitly converts it to char maybe

McLion

@raven-worx
Thanks. I implemented it the (longer) way it works.

Anybody an idea about the Debug() question?
Sorry I put 2 issues in one thread ;-)

raven-worx

@McLion said in QByteArray increment, and qDebug() benaviour:

Anybody an idea about the Debug() question?

try passing char/char* to qDebug(), instead of the Qt classes.

McLion

@raven-worx

Didn't help ... never mind.

JKSH

@McLion said in QByteArray increment, and qDebug() benaviour:

I regularly use this in C without issues.

How do you print it in C? Note:

printf("%d\n", '1') prints "49"
printf("%c\n", 49) prints "1"

Similarly, QDebug formats char values using %c and formats int values using %d.

Note also, when you add/subtract char values, the compiler (at least MSVC 2013 and GCC 4.8) implicitly casts them to int. Proof:

// Input:
#include <typeinfo>
...
qDebug() << typeid('0' + 1).name();
qDebug() << typeid('0' - '0').name();

// Output (on MSVC 2013 and GCC 4.8):
int
int

If I add '0' it should be 0x31 which is '1', but it outputs 49

QDebug receives (int)0x31, so it prints "49".

If I do +'0'-48 - which actually seems very stupid, does it - it shows a correct '1'

Your code is 1 + '0' - 48 == 1 + 48 - 48 == 1.

QDebug receives a (int)0x01, so it prints "1".

McLion

@JKSH
Thanks a lot for this detailed explanation ... which explains a lot.
The value in question is from a QByteArray and therefore is a char.
When casting to int instead of doing the stupid +'0'-48, everything works as expected.

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QByteArray increment, and qDebug() benaviour