[英]How to make unique pointers to polymorphic classes using the ternary operator?
I'm trying to set a variable using the ternary operator.我正在尝试使用三元运算符设置变量。 However, the compiler is complaining about incompatible types.但是,编译器抱怨不兼容的类型。 I'm sure there is a way to do this.我确信有办法做到这一点。 I have tried static casting to the base class, but I haven't been able to get the correct syntax.我已经尝试将 static 转换为基础 class,但我无法获得正确的语法。
#include <iostream>
#include <memory>
struct A
{
virtual ~A() = default;
virtual void test() {std::cout << "A" << std::endl;}
};
struct B: public A
{
void test() final {std::cout << "B" << std::endl;}
};
struct C: public A
{
void test() final {std::cout << "C" << std::endl;}
};
int main()
{
bool t = true;
// Try to cast try a base unique class ptr. Maybe use static_cast??
std::unique_ptr<A> aptr = t ? std::make_unique<B>(): std::make_unique<C>();
aptr->test();
}
Return value of ternary expression is the common type of both expressions (in fact std::common_type
might use ternary operator as part of its implementation:-) ).三元表达式的返回值是两个表达式的共同类型(实际上std::common_type
可能使用三元运算符作为其实现的一部分:-) )。
Whereas std::unique_ptr<B>
and std::unique_ptr<C>
are unrelated types,而std::unique_ptr<B>
和std::unique_ptr<C>
是不相关的类型,
both std::unique_ptr<B>
and std::unique_ptr<C>
are convertible to std::unique_ptr<A>
, so converting one explicitly would be enough: std::unique_ptr<B>
和std::unique_ptr<C>
都可以转换为std::unique_ptr<A>
,因此显式转换一个就足够了:
auto aptr = t ? std::unique_ptr<A>{std::make_unique<B>()}
: std::make_unique<C>();
Try this:尝试这个:
std::unique_ptr<A> aptr = t ? std::unique_ptr<A>(new B()): std::unique_ptr<A>(new C());
You can do a direct call to the constructor:您可以直接调用构造函数:
std::unique_ptr<A> aptr(t ? std::make_unique<B>(): std::make_unique<C>());
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