QVariant pointer not respond value

Kutyus

Hi!

I have a QVariant pointer:

QVariant *param;

I passing to an integer value:

int p = 12;

param = (QVariant*)(&p);

But i do not get the value, i get always 0:

param->toInt();

The gdb write this:

(gdb) print *param
$7 = {d = {data = {c = 12 '\f', uc = 12 '\f', s = 12, sc = 12 '\f', us = 12, i = 12, u = 12, l = 12, ul = 12, b = 12, d = 5.1185933671348506e-309, f = 1.68155816e-44, real = 5.1185933671348506e-309, ll = 1036014831271948,
      ull = 1036014831271948, o = 0xc, ptr = 0xc, shared = 0xc}, type = 0, is_shared = 0, is_null = 0}}

What do I get rid of?

Thank you for answers.

raven-worx

@Kutyus said in QVariant pointer not respond value:

param->toInt();

honestly i do not know why your code doesn't crash o.O
You are casting the stack-pointer address of p to a QVariant pointer and later you access this invalid pointer.

I see what you are trying to do and it should rather look like this:

param = new QVariant(p);

Venkatesh V

Hi,
@Kutyus said in QVariant pointer not respond value:

param = (QVariant*)(&p);

instead you do like this, try this
param =QVariant::fromValue(&p);

Kutyus

I did not write down everything, of course there is a

param = new QVariant();

line.

Kutyus

@Venkatesh-V
It is wrong, param is QVariant*, not QVariant.

raven-worx

@Venkatesh-V said in QVariant pointer not respond value:

param =QVariant::fromValue(&p);

this also results in an invalid pointer! DO NOT use stack-pointer addresses for such cases.

@Kutyus said in QVariant pointer not respond value:

I did not write down everything, of course there is a
param = new QVariant();

line.

Still you want to use QVariant implicit conversion constructor / assignment operator. So pass the int value not it's pointer address!

kshegunov

@raven-worx said in QVariant pointer not respond value:

honestly i do not know why your code doesn't crash o.O

Nothing to crash, but it's pure "luck". ;)

Kutyus

I modified the concept, and use QVariant, not QVariant*, it's working.