问题描述
在 cppreference 的 std::enable_if 的语言参考中 包含以下注释
In the language reference of std::enable_if at cppreference the following note is included
注意事项
一个常见的错误是声明两个不同的函数模板仅在它们的默认模板参数中. 这是非法的，因为默认模板参数不是函数模板的一部分签名，并声明两个不同的函数模板相同的签名是非法的.
A common mistake is to declare two function templates that differ only in their default template arguments. This is illegal because default template arguments are not part of function template's signature, and declaring two different function templates with the same signature is illegal.
在下面示例中的模板函数中，在我看来，这种情况发生了.即，两个模板函数 onlyForDerivedObjects(...) 似乎(对我而言)仅在默认模板参数上有所不同.我意识到我在这里遗漏了一些东西，希望有人可以向我解释这一点，或者为我指明方向，我可以找到自己的顿悟.
In the template functions in example below, it seems to me that this situation occurs. I.e., the two template functions onlyForDerivedObjects(...) seem (to me) to differ only by their default template arguments. I realize I am missing something here, and hopefully someone can explain this to me, or point me in the direction to where I might find an epiphany for myself.
- 问题: W.r.t.上面的引用，为什么下面的例子编译和运行良好:当我认为它产生具有两个模板的情况时，我是否错误分类了下面模板函数中的
typename std::enable_if ...部分仅在默认模板参数方面不同的函数?
- Question: W.r.t. the quote above, why do the example below compile and run fine: do I misclassify the
typename std::enable_if ...part in the template functions below when I consider it to yield a situation with two template functions which differ only in their default template argument?
基类和派生类:
class BaseA
{
public:
int getInt() const { return 21; };
};
class DerivedA : public BaseA {};
class BaseB
{
public:
int getAnotherInt() const { return 33; };
};
class DerivedB : public BaseB {};
具有以下模板函数
/* template functions that, seemingly, only differ in their
default template arguments? */
template< class T,
typename std::enable_if<std::is_base_of<BaseA, T>::value>::type* = nullptr >
int onlyForDerivedObjects(const T& obj)
{
return 2*obj.getInt();
}
template< class T,
typename std::enable_if<std::is_base_of<BaseB, T>::value>::type* = nullptr >
int onlyForDerivedObjects(const T& obj)
{
return 3*obj.getAnotherInt();
}
编译并运行良好(g++ -Wall -std=c++11 ..., g++ 4.9.3)
compiles and runs fine (g++ -Wall -std=c++11 ..., g++ 4.9.3)
#include <iostream>
#include <type_traits>
/* ... classes and template functions as above */
/* template argument deduction seems to work fine */
int main()
{
DerivedA* objA = new DerivedA();
DerivedB* objB = new DerivedB();
std::cout << onlyForDerivedObjects(*objA) << std::endl; // 42
std::cout << onlyForDerivedObjects(*objB) << std::endl; // 99
return 0;
}
推荐答案
注意事项
一个常见的错误是声明两个仅在默认模板参数上不同的函数模板.这是非法的，因为默认模板参数不是函数模板签名的一部分，并且声明具有相同签名的两个不同函数模板是非法的.
A common mistake is to declare two function templates that differ only in their default template arguments. This is illegal because default template arguments are not part of function template's signature, and declaring two different function templates with the same signature is illegal.
您的函数不仅在默认模板参数方面有所不同，而且在模板参数方面也有所不同，因此具有不同的签名.
Your functions don't differ only in their default template arguments, they differ in their template parameters, so have different signatures.
在这两种情况下，默认模板参数都是nullptr，但第二个模板参数在每种情况下都不同.
In both cases the default template argument is nullptr, but the second template parameter is different in each case.
这篇关于使用 std::enable_if 作为模板时的默认模板参数.参数:为什么两个模板函数只在 enable_if 参数上不同?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，WP2