[Development] Setters: Passing by value or const reference?

[Thiago Macieira]

On quarta-feira, 25 de abril de 2012 13.57.31, Alberto Mardegan wrote:
> I understand why there wouldn't be a copy in the implementation of
> setText, but why wouldn't calling foo->setText() produce a copy?
> How can the compiler know that it must not create a copy when calling
> Foo::setText(QString text) unless we are inlining it?

Ah, you're missing the trick!

Olivier has just updated https://codereview.qt-project.org/24144 "Implement 
the move consrtuctor for containers" and most other tool classes already have 
move constructors.

If you have:

	void setter(QString s);

and you call it with:

	QString foo("foo");
	setter(foo);		// 1
	setter(QString("bar")); //2

In both cases, the compiler must call a QString constructor because you're 
passing by value. In the first case, since foo is an lvalue, it chooses

	QString::QString(const QString &)
which does the normal, cheap reference counting up.

In the second case, since it's an rvalue, it chooses

	QString::QString(QString &&)
which is the move constructor: it doesn't reference count up, it simply 
"steals" the d pointer from the parameter and resets the parameter to 
QString()

Sprinkle a few Q_ASSUME() in some places and the compiler will know that the 
QString destructor for the temporary is empty.

-- 
Thiago Macieira - thiago.macieira (AT) intel.com
  Software Architect - Intel Open Source Technology Center
     Intel Sweden AB - Registration Number: 556189-6027
     Knarrarnäsgatan 15, 164 40 Kista, Stockholm, Sweden
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