C++ 为什么使用 enable_if 编译错误
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Why compile error with enable_if
提问by Leonid Volnitsky
Why this does not compile with gcc48 and clang32?
为什么这不能用 gcc48 和 clang32 编译？
#include <type_traits>
template <int N>
struct S {
template<class T>
typename std::enable_if<N==1, int>::type
f(T t) {return 1;};
template<class T>
typename std::enable_if<N!=1, int>::type
f(T t) {return 2;};
};
int main() {
S<1> s1;
return s1.f(99);
}
GCC error:
海湾合作委员会错误：
/home/lvv/p/sto/test/t.cc:12:2: error: no type named ‘type' in ‘struct enable_if<false, int>'
f(T t) {return 2;};
^
CLANG error:
CLANG 错误：
/home/lvv/p/sto/test/t.cc:11:26: error: no type named 'type' in 'std::enable_if<false, int>'; 'enable_if' cannot be used to
disable this declaration
typename std::enable_if<N!=1, int>::type
^~~~
/home/lvv/p/sto/test/t.cc:16:7: note: in instantiation of template class 'S<1>' requested here
S<1> s1;
^
EDIT - SOLUTION
编辑 - 解决方案
I've accepted answer from Charles Salvia, but for practical reasons I was not able to use workaround that was proposed (specialize on N). I found other workaround which works for me. Make enable_ifdepend on T:
我已经接受了 Charles Salvia 的回答，但出于实际原因，我无法使用提出的解决方法（专门针对 N）。我找到了其他对我有用的解决方法。使enable_if依赖于T：
typename std::enable_if<(sizeof(T),N==1), int>::type
采纳答案by Charles Salvia
Because you use enable_ifwithout using the template parameter Tin your function templates. If you want to specialize for when the struct Shas a certain template parameter value N, you'll need to use class template specialization.
因为您在函数模板中使用enable_if而不使用模板参数T。如果您想专门用于结构S具有特定模板参数值的情况N，则需要使用类模板专门化。
template <int N, class Enable = void>
struct S { };
template <int N>
struct S<N, typename std::enable_if<N == 1>::type>
{
....
};
回答by foxfireee
Well, I am not sure what you wanted to do, but maybe this code will help:
好吧，我不确定你想做什么，但也许这段代码会有所帮助：
#include <iostream>
template <int N>
struct S {
template<class T=int>
typename std::enable_if<N==1, T>::type
f(T t) {return 1;}
template<class T=int>
typename std::enable_if<N!=1, T>::type
f(T t) {return 2;}
};
int main()
{
S<1> s1;
S<2> s2;
std::cout << s1.f(99) << " " << std::endl << s2.f(5);
}
This prints 1 and 2.
这将打印 1 和 2。
回答by Paul Fultz II
Use a default boolean template parameter, like this:
使用默认的布尔模板参数，如下所示：
template <int N>
struct S {
template<class T, bool EnableBool=true>
typename std::enable_if<N==1 && EnableBool, int>::type
f(T t) {return 1;};
template<class T, bool EnableBool=true>
typename std::enable_if<N!=1 && EnableBool, int>::type
f(T t) {return 2;};
};
回答by Anthony
To get std::enable_ifto work like this, you are relying on SFINAE. Unfortunately, at the point where you declare
要std::enable_if像这样开始工作，您需要依赖 SFINAE。不幸的是，在您声明时
S<1> s1;
it will instantiate all of S<1>'s member declarations. SFINAE will only come into play at this point if S<1>were an ill-formed construct. It is not. Unfortunately, it contains a function which is invalid, thus the instantiation of S<>is invalid.
它将实例化所有S<1>的成员声明。如果 SFINAES<1>是一个格式错误的结构，那么此时它只会发挥作用。它不是。不幸的是，它包含一个无效的函数，因此实例化S<>是无效的。
For things like this, I might defer to a seperate template struct:
对于这样的事情，我可能会遵循一个单独的模板结构：
template <bool B>
struct f_functor {
template <typename T>
static int f(T t) { return 1; }
};
template <>
struct f_functor<false> {
template <typename T>
static int f(T t) { return 2; }
};
template <int N>
struct S {
template<class T>
typename int f(T t) { return f_functor<N==1>::f(t); }
};
回答by Jan Herrmann
For this case you could think about not using enable_if at all. It is posible to simply specialise f:
对于这种情况，您可以考虑根本不使用 enable_if。可以简单地特化 f：
template <int N>
struct S {
template<class T> int f(T t);
};
template<int N>
template<class T>
int S<N>::f(T t) { return 2; }
template<>
template<class T>
int S<1>::f(T t) { return 1; }
int main() {
S<1> s1;
return s1.f(99);
}
