dynamic_pointer_cast
描述 (Description)
它返回一个正确类型的sp副本，其存储的指针从U *动态地转换为T *。
声明 (Declaration)
以下是std :: dynamic_pointer_cast的声明。
template <class T, class U>
shared_ptr<T> dynamic_pointer_cast (const shared_ptr<U>& sp) noexcept;
C++11
template <class T, class U>
shared_ptr<T> dynamic_pointer_cast (const shared_ptr<U>& sp) noexcept;
参数 (Parameters)
sp - 它是一个共享指针。
返回值 (Return Value)
它返回一个正确类型的sp副本，其存储的指针从U *动态地转换为T *。
异常 (Exceptions)
noexcep - 它不会抛出任何异常。
例子 (Example)
在下面的示例中解释了std :: dynamic_pointer_cast。
#include <iostream>
#include <memory>
struct A {
static const char* static_type;
const char* dynamic_type;
A() { dynamic_type = static_type; }
};
struct B: A {
static const char* static_type;
B() { dynamic_type = static_type; }
};
const char* A::static_type = "sample text A";
const char* B::static_type = "sample text B";
int main () {
std::shared_ptr<A> foo;
std::shared_ptr<B> bar;
bar = std::make_shared<B>();
foo = std::dynamic_pointer_cast<A>(bar);
std::cout << "foo's static type: " << foo->static_type << '\n';
std::cout << "foo's dynamic type: " << foo->dynamic_type << '\n';
std::cout << "bar's static type: " << bar->static_type << '\n';
std::cout << "bar's dynamic type: " << bar->dynamic_type << '\n';
return 0;
}
让我们编译并运行上面的程序，这将产生以下结果 -
foo's static type: sample text A
foo's dynamic type: sample text B
bar's static type: sample text B
bar's dynamic type: sample text B
