static_cast和引用指针 [英] static_cast and reference to pointers
问题描述
任何人都可以告诉我为什么这不编译:
Can anyone tell me why this doesn't compile:
struct A { };
struct B : public A { };
int main()
{
B b;
A* a = &b;
B* &b1 = static_cast<B*&>(a);
return 0;
}
现在,如果将静态转型替换为:
Now, if you replace the static cast with:
B* b1 = static_cast<B*>(a);
那么它会编译。
strong> Edit:显然,编译器将 A *
和 B *
视为独立类型,否则这将工作。
It is obvious that the compiler treats A*
and B*
as independent types, otherwise this would work. The question is more about why is that desirable?
推荐答案
B
是源自 A
,但 B *
不是从 A *
。
指向 B
的指针不是指向 A
的指针,它只能是
转换为一。但是类型保持不变(并且
转换可以,并且经常会改变
指针的值)。 B *&
只能指向 B *
,而不能指向任何其他
指针类型。
B
is derived from A
, but B*
isn't derived from A*
.
A pointer to a B
is not a pointer to an A
, it can only be
converted to one. But the types remain distinct (and the
conversion can, and often will, change the value of the
pointer). A B*&
can only refer to a B*
, not to any other
pointer type.
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