为什么使用enable_if编译错误 [英] Why compile error with enable_if

问题描述

为什么这不能与gcc48和clang32一起编译?

Why this does not compile with gcc48 and clang32?

#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;
             ^

编辑-解决方案

我已经接受了Charles Salvia的回答，但是出于实际原因，我无法使用建议的解决方法(专门针对N).我找到了其他对我有效的解决方法.使enable_if依赖于T:

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_if depend on T:

typename std::enable_if<(sizeof(T),N==1), int>::type

推荐答案

因为您使用enable_if而不在功能模板中使用模板参数T.如果要在结构S具有特定模板参数值N时进行专业化，则需要使用类模板专业化.

Because you use enable_if without using the template parameter T in your function templates. If you want to specialize for when the struct S has a certain template parameter value N, you'll need to use class template specialization.

template <int N, class Enable = void> 
struct S {  };

template <int N>
struct S<N, typename std::enable_if<N == 1>::type>
{
  ....
};

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