Qt - QList const正确性 [英] Qt - QList const correctness

问题描述

A QList< T *> 不能容易地是const正确的。考虑函数

A QList<T *> can't easily be const-correct. Consider the function

void f(QList<T *> list)
{
    list[0]->constFunction();
}

我可以将f更改为

void f(QList<const T *> list)

$ b b

但我不能做

but then I can't do

f(QList<T *>());    //Compile error

因为编译器不能隐式转换 QList ; T *> 到 QList< const T *> 。然而，我可以明确地重新解释QList如下：

anymore, since the compiler can't implicitely cast QList<T *> to QList<const T *>. However, I can explicitely reinterpret-cast the QList as follows:

template <typename T> inline QList<const T *> &constList(const QList<T *> &list)
{
    return (QList<const T *> &)list;
}

这使我可以使用 constList 模板函数将任何 QList 转换为 QList 如在

This enables me to use the constList template function to cast any QList<T *> into a QList<const T *>, as in

f(constList(QList<T *>()));

它似乎工作正常，但它是否真的可以安全吗？

and it seems to work fine, but is it actually safe to do this?

推荐答案

您正在考虑的投射功能…

The casting function you're considering, …

template< class T >
inline QList<T const*>& constList( QList<T*> const& list )
{
    return reinterpret_cast< QList<T const*>& >(
        const_cast< QList<T*>& >( list )
        );
}

…可能是实用的（可能 QList 不会根据元素类型的 const -ness改变其对象表示）但它可以打破 const 正确。

… may be practical (probably QList does not change its object representation depending on the const-ness of the element type), but it can break const correctness.

c $ c> const - 列表本身的值不是 const 正确：它允许你改变一个最初的 const

First, because casting away the const-ness of the list itself is not const correct: it allows you to change an originally const list.

但是即使删除了形式参数 const ，像…

But even if that formal argument const is removed, like …

template< class T >
inline QList<T const*>& constList( QList<T*>& list )
{
    return reinterpret_cast< QList<T const*>& >(
        list
        );
}

…仍然有一个 const 正确性问题。

… there is still a const correctness problem.

原因是列表构成一个附加级别的间接，你的函数本身不是 const 。因此，在使用你的函数获得一个引用到指向指向 - const 元素的列表之后，你可以在那个列表中存储一个指向真正 const 。然后，您可以使用原始列表修改真的 const 项目 bang 。

The reason is that the list constitutes an additional level of indirection, and with your function is not itself const. Thus, after using your function to get a reference to the list with alleged pointer-to-const elements, you can store in that list a pointer to something that is really const. And then you can use the original list to modify that really const item, bang.

这与没有隐含从 T ** 转换为 T const ** 。

It's the same reason that there is no implicit conversion from T** to T const**.

没有这样的问题你可以做的是，已经有 const 指向对象的指针列表，使那些指向对象 const ：

What you can do without such problems is, with an already const list of pointers to objects, make those pointed to objects const:

template< class T >
inline QList<T const*> const& constList( QList<T*> const& list )
{
    return reinterpret_cast< QList<T const*> const& >(
        list
        );
}

正式还有 reinterpret_cast 作为一个潜在的问题，但任何专门化的表示的元素的元素的const的元素的任何专门化的代表可能应该得到的任何他们得到的$ QList 。 ： - ）

Formally there's still the reinterpret_cast as a potential problem, but anyone specializating the representation of QList on constness of elements would presumably deserve whatever they got. :-)

干杯& hth。，

Cheers & hth.,

这篇关于Qt - QList const正确性的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！