How to implement QGLFormat with QGLContext

What I'd like to know is how pass my QGLFormat object to QGLContext?

When I try to do this:

Code:

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QGLFormat mFormat = QGLFormat( QGL::DepthBuffer | QGL::AlphaChannel );
 
QGLContext mContext = QGLContext( mFormat );

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I get this error:

Code:

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error: 'QGLContext::QGLContext(const QGLContext&)' is private

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Using Qt Creator with Windows 7.

What is wrong here?

Change

Code:

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QGLContext mContext = QGLContext( mFormat );

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to

Code:

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QGLContext mContext( mFormat );

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I think compiler "thinks" you want to use copy-constructor, which is private. But in fact the copy-constructor will not be used in this case. Look here:

Code:

---

#include <iostream>
 
class T
{
public:
		T(){
			std::cout << "T()\n";
		}
		T(const T&){
			std::cout << "T(const T&)\n";
		}
		T(int){
			std::cout << "T(int)\n";
		}
		T& operator=(const T& t){
			std::cout << "T& operator\n";
		}
};
 
int main(){
	T t = T(10);
	return 10;
}

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This should print T(int), which means the copy-constructor is not used. But this:

Code:

---

#include <iostream>
 
class T
{
public:
		T(){
			std::cout << "T()\n";
		}
		T(int){
			std::cout << "T(int)\n";
		}
		T& operator=(const T& t){
			std::cout << "T& operator\n";
		}
private:
		T(const T&){
			std::cout << "T(const T&)\n";
		}
};
 
int main(){
	T t = T(10);
	return 10;
}

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does not compile:

Quote:

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test.cpp: In function 'int main()':
test.cpp:54:3: error: 'T::T(const T&)' is private
test.cpp:60:12: error: within this context

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(tested with g++ 4.5.2)