static_cast(*this) to a base class create a temporary copy.
class Window { // base class
public:
virtual void onResize() { ... } // base onResize impl
...
}; class SpecialWindow: public Window { // derived class
public:
virtual void onResize() { // derived onResize impl;
static_cast<Window>(*this).onResize(); // cast *this to Window,
// then call its onResize;
// this doesn't work!
... // do SpecialWindow-
} // specific stuff ... };
Effective C++: What you might not expect is that it does not invoke that function on the current object! Instead, the cast creates a new, temporary copy of the base class part of *this, then invokes onResize on the copy!
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Contrast:
static_cast<Window>(*this)
with:
static_cast<Window&>(*this)
One calls the copy constructor, the other does not.
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Because you are casting actual object not a pointer or reference. It's just the same with casting double
to int
creates new int
- not reusing the part of double
.
double类型转换为int型会创建一个新的int型变量?
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上面的句子static_cast<Window>(*this).onResize();千万别改成下面这样,那样会更悲惨!
virtual void onResize()
{
static_cast<Window*>(this)->onResize(); //调用基类的onResize()
derived_ = 2;
std::cout <<"S" << " base_=" << base_ << ",derived=" << derived_ << std::endl;
}
实际上,可能只有C的高手初转C++,对C++对象的内存模型还不很清楚的情况下才会写成这样的代码。
对一个指针,不论我们怎样强制转换,它指向的那段内存的内容并没有改变。
函数把本类对象的指针强制转换为基类对象的指针,只意味着“通过这个指针只能访问基类的成员”了,而对象的内容并没有改变。onResize是虚函数,调用它是先通过对象中指向虚表的指针找到虚表,然后在虚表中找到onResize函数的指针,最后通过函数指针调用函数。 this指针强制转换后,内存没有改变,所以指向虚表的指针没有改变,所以它找到的虚表仍是派生类的虚表,自然找到的函数指针仍是派生类的onResize函数的指针,所以这里就成了一个无穷递归调用,结果就是消耗完栈空间。
将函数修改为这样:你就会看到不断的输出ReCall.
virtual void onResize()
{
std::cout<<"ReCall"<<endl;
static_cast<Window*>(this)->onResize(); //调用基类的onResize()
derived_ = 2;
std::cout <<"S" << " base_=" << base_ << ",derived=" << derived_ << std::endl;
}