[SOLVED] How to copy QImage into a part of another QImage
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Hi all,
I have 2 QImages and I want to copy the second one into the first one but in a specific rectangle.
Basically I want to do this:
QImage1(x1, y1, w1, h1)= QImage2->copy(x2 ,y2, w2, h2);
but QImage1 doesn't have any property to say to copy into a specific location. Should I use QPainter?
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use QPainter :D
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[quote author="Code_ReaQtor" date="1375092283"]use QPainter :D[/quote]
That's good :D
Actually I started trying QPainter after I post my message here.
I wrote the following code but it crashes in pntToOut->drawPicture.@
QImage *tmpImage=new QImage(origW, origH, orgImg.format());
//Do some stuff in tmpImage
QPainter *pntToOut = new QPainter(&orgImg);
pntToOut->drawPicture(x,y, (const QPicture &)tmpImage);
@ -
why do you cast?!
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QPainter p(&orgImage);
p.drawImage(x,y, tmpImage);
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[quote author="raven-worx" date="1375093084"]
why do you cast?!
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QPainter p(&orgImage);
p.drawImage(x,y, tmpImage);
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[/quote]Because I got the following error if I don't cast:
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'void QPainter::drawImage(const QRectF &,const QImage &)' : cannot convert parameter 2 from 'QImage *' to 'const QImage &'
Reason: cannot convert from 'QImage *' to 'const QImage'
No constructor could take the source type, or constructor overload resolution was ambiguous
@ -
Hi,
No real need to allocate a new image
@
QImage tmpImage(origW, origH, orgImg.format());
//Do some stuff in tmpImage
QPainter *pntToOut = new QPainter(orgImg);
pntToOut->drawPicture(x,y, tmpImage);
@ -
[quote author="SGaist" date="1375093649"]Hi,
No real need to allocate a new image
@
QImage tmpImage(origW, origH, orgImg.format());
//Do some stuff in tmpImage
QPainter *pntToOut = new QPainter(orgImg);
pntToOut->drawPicture(x,y, tmpImage);
@[/quote]Aha, that solved it. thanks a lot.
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[quote author="dolevo" date="1375093316"]
Because I got the following error if I don't cast:
[/quote]
The error says that the method expects an value rather than a pointer.
So you either allocate it on the stack (like SGaist suggested) or you pass the value of the pointer (if you need to stick to pointer):
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p.drawImage(x,y, *tmpImage);
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[quote author="raven-worx" date="1375094384"][quote author="dolevo" date="1375093316"]
Because I got the following error if I don't cast:
[/quote]
The error says that the method expects an value rather than a pointer.
So you either allocate it on the stack (like SGaist suggested) or you pass the value of the pointer (if you need to stick to pointer):
@
p.drawImage(x,y, *tmpImage);
@[/quote]
I understand. Thank you very much for your detailed explanation. I appreciate it.
