[英]C++ template function for derived class with std::is_base_of
I've got problem with creating function that for given type, if it's derived from other one do something and for all other cases do something other.我在为给定类型创建函数时遇到问题,如果它派生自其他类型,则执行某些操作,而对于所有其他情况,则执行其他操作。 My code:
我的代码:
class BaseClass {};
class DerivedClass : public BaseClass {};
template <typename T>
void Function(typename std::enable_if<std::is_base_of<BaseClass, T>::value, T>::type && arg) {
std::cout << "Proper";
}
template <typename T>
void Function(T && arg) {
std::cout << "Improper";
}
void test() {
Function(DerivedClass{});
}
For class DeriviedClass
and other based on BaseClass
I'd like to call function couting Proper
, but it couts Improper
.对于类
DeriviedClass
和其他基于BaseClass
我想调用函数couts Proper
,但它couts Improper
。 Any suggestions?有什么建议?
As mentioned in the comments to the question, SFINAE expressions won't work the way you did it.正如对问题的评论中提到的,SFINAE 表达式不会像你那样工作。
It should be instead something like this:它应该是这样的:
template <typename T>
typename std::enable_if<std::is_base_of<BaseClass, T>::value>::type
Function(T && arg) {
std::cout << "Proper" << std::endl;
}
template <typename T>
typename std::enable_if<not std::is_base_of<BaseClass, T>::value>::type
Function(T && arg) {
std::cout << "Improper" << std::endl;
}
SFINAE expressions will enable or disable Function
depending on the fact that BaseClass
is base of T
. SFINAE 表达式将根据
BaseClass
是T
基础这一事实启用或禁用Function
。 Return type is void
in both cases, for it's the default type for std::enable_it
if you don't define it.在这两种情况下,返回类型都是
void
,因为如果您没有定义它,它就是std::enable_it
的默认类型。
See it on coliru .在coliru上看到它。
Other valid alternatives exist and some of them have been mentioned in other answers.存在其他有效的替代方案,其中一些已在其他答案中提到。
#include <typeinfo>
#include <iostream>
class BaseClass {};
class DerivedClass : public BaseClass {};
class OtherClass {};
template <typename T,typename = typename std::enable_if<std::is_base_of<BaseClass, T>::value, T>::type>
void Function(T && arg)
{
std::cout << "Proper" << std::endl;
}
void Function(...)
{
std::cout << "Improper"<< std::endl;
}
int main()
{
Function(DerivedClass{});
Function(BaseClass{});
Function(OtherClass{});
}
template <typename T>
auto Function(T && arg) -> typename std::enable_if<std::is_base_of<BaseClass, T>::value>::type
{
std::cout << "Proper";
}
template <typename T>
auto Function(T && arg) -> typename std::enable_if<!std::is_base_of<BaseClass, T>::value>::type
{
std::cout << "Improper";
}
C++11+ : C++11+:
#include <type_traits> // for is_base_of<>
class Base {};
class Derived : public Base {};
class NotDerived {};
template<typename Class>
void foo(const Class& cls)
{
static_assert(is_base_of<Base, Class>::value, "Class doesn't inherit from Base!");
// The codes...
}
int main()
{
foo(Derived()); // OK!
foo(NotDerived()); // Error!
return 0;
}
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