[Solved]what does this syntax mean in c++?

whitesong

I want to know what this syntax says:

@(void) new ClassName();@

the syntax above is used in this snip code for populate a widget.

@ void populateTree(QTreeWidget *treeWidget)
{
QStringList labels;
labels << QObject::tr("Terms") << QObject::tr("Pages");

    treeWidget->setHeaderLabels(labels);
    treeWidget->header()->setResizeMode(QHeaderView::Stretch);
    treeWidget->setWindowTitle(QObject::tr("XML Stream Writer"));
    treeWidget->show();

    (void) new QTreeWidgetItem(treeWidget,
            QStringList() << "sidebearings" << "10, 34-35, 307-308");
    QTreeWidgetItem *subterm = new QTreeWidgetItem(treeWidget,
            QStringList() << "subtraction");
    (void) new QTreeWidgetItem(subterm,
            QStringList() << "of pictures" << "115, 244");
    (void) new QTreeWidgetItem(subterm,
            QStringList() << "of vectors" << "9");
}@

coyotte508

The (void) doesn't do anything, it just removes some warnings for some IDEs to tell it's normal the result of the "new" isn't stored in any variable.

whitesong

thanks. so in some IDEs "(void)" will be needed

utcenter

It is not needed in any of the popular IDEs and compilers.

JKSH

[quote author="whitesong" date="1367141103"]thanks. so in some IDEs "(void)" will be needed[/quote]It's not really needed, but it's nice to get rid of compiler warnings.

In this code below, your compiler might warn you that x and y are unused. :
@
void myFunction(int x)
{
int y = 1;
return; // Produces a warning that x and y are unused
}
@

But, calling (void) tells the compiler to "do nothing" with the variables, so it won't complain:

@
void myFunction(int x)
{
int y = 1;

// Prevents warnings
(void) x;
(void) y;

return;

}
@

Qt has its own macro to do this, instead of (void). The code below has exactly the same effect as the code above:
@
void myFunction(int x)
{
int y = 1;

// Prevents warnings
Q_UNUSED(x);
Q_UNUSED(y);

return;

}
@

whitesong

Oh thanks again everyone.
I really understand what is going on with your explanation .

whitesong

Oh thanks again everyone.
I really understand what is going on with your explanation.