我如何理解这种静态铸造？ [英] How do I understand this static casting ?

问题描述

我在理解以下代码行时遇到问题：

I have a problem with understanding the following line of code:

static_cast<void(QButtonGroup::*)(QAbstractButton *, bool)>(&QButtonGroup::m_buttonGroup)

我的尝试：



我搜索了static_cast，并了解它，我知道在这段代码中我们试图转换对象

What I have tried:

I searched for the static_cast, and understand it, i know that in this code we are trying to convert the object

(&QButtonGroup::m_buttonGroup)

到另一种类型：

to another type:

<void(QButtonGroup::*)(QAbstractButton *, bool)>

但我不明白上一行。我们的意思是：

But i do not understand the previous line.what do we mean by:

void(QButtonGroup::*)(QAbstractButton *, bool)

这应该是一个类型，但我无法得到它？

提前谢谢。

this supposed to be a type, but i can not get it?
Thanks in advance.

推荐答案

static_cast<void(QButtonGroup::*)(QAbstractButton *, bool)>(&QButtonGroup::m_buttonGroup)

Star从错误的一端看看括号中的任何内容：

Starting from the wrong end look at whatever is in the parentheses:

&QButtonGroup::m_buttonGroup

m_buttonGroup 是一个静态对象，位于 QButtonGroup 类中。 & 表示使用此对象的地址。这就是要投射的原始项目，以及尖括号中的任何内容（<> ）。尖括号包含：

m_buttonGroup is a static object, in the QButtonGroup class. And the & means use the address of this object. So that is the original item that is to be cast, to whatever is in the angle brackets (< >). The angle brackets contain:

void(QButtonGroup::*)(QAbstractButton *, bool)

哪个是函数指针定义。 （QButtonGroup :: *）部分表示这是指向 QButtonGroup 类中某些内容的指针。围绕它的括号告诉我们它是一个函数而不是数据指针。下面的括号包含要传递给函数的参数的定义，前面的 void 部分告诉我们该函数没有返回类型。



如果没有更多背景信息，就不可能多说。

Which is a function pointer definition. The (QButtonGroup::*) part says this is a pointer to something in the QButtonGroup class. The parentheses around this tell us it's a function rather than a data pointer. The following parentheses contain the definition of the parameters to be passed to the function, and the preceding void part tells us that the function does not have a return type.

Without more context it's not possible to say more.

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