C ++模板sfinae错误 [英] C++ template sfinae error

问题描述

我下面有一些代码，我希望具有不同专业的TestEnableIf具有不同的打印功能，但是它没有按计划进行，出现以下错误.

I have below code that I want TestEnableIf with different specialization to have different print function, but it didn't work out as planned, with error as below.

struct myStruct;
struct notMyStruct;

template<typename T> struct TestEnableIf
{
        template<typename boost::enable_if< boost::is_same<myStruct, T>, int >::type = 0> void print()
        {
            std::cout << "this is my struct" << std::endl;
        }

        template<typename boost::enable_if< boost::is_same<notMyStruct, T>, int>::type=0> void print()
        {
            std::cout << "This is not my struct" << std::endl;
        }


};

static void testIsSame()
{

    TestEnableIf<myStruct> testEnable;
     testEnable.print();
     TestEnableIf<notMyStruct> testNotEnable;
     testNotEnable.print();
}

../src/testBoostGeneric.h:39:90:错误:在其中没有名为"type"的类型 ‘struct boost :: enable_if，int>’ 模板， int> :: type = 0> void print() ^ ../src/testBoostGeneric.h:在"struct"的实例中 TestEnableIf":../src/testBoostGeneric.h:53:29:
从这里需要../src/testBoostGeneric.h:34:90:错误:无类型 在'struct boost :: enable_if，int>'模板中被命名为'type'，int> :: type = 0> void print()

../src/testBoostGeneric.h:39:90: error: no type named ‘type’ in ‘struct boost::enable_if, int>’ template, int>::type=0> void print() ^ ../src/testBoostGeneric.h: In instantiation of ‘struct TestEnableIf’: ../src/testBoostGeneric.h:53:29:
required from here ../src/testBoostGeneric.h:34:90: error: no type named ‘type’ in ‘struct boost::enable_if, int>’ template, int >::type = 0> void print()

我不明白的是，sfinae表示专业化失败不是错误，那么编译器为什么要抱怨失败?

What I don't understand is, should sfinae mean specialization failure is not an error, then why should the compiler complain about the failure?

推荐答案

SFINAE(顺便说一下，S是替代"，而不是专业化")不适用于类类型参数.解决该问题的一种简单方法是使用另一个模板参数:

SFINAE (the S is "substitution", not "specialization", by the way) for member functions does not work this way with class type parameters. A simple way to get around that is to use another template parameter:

template<
    typename TCopy = T
    typename boost::enable_if< boost::is_same<myStruct, TCopy>, int >::type = 0
> void print()
{
    std::cout << "this is my struct" << std::endl;
}

我从 STL的CppCon谈话中撤了下来. TL; DW:使用类的T类型参数时，编译器在实例化类模板时知道该类型是什么，并会在该时间点检查type成员.使用额外的TCopy类型参数位于函数本地时，编译器无法确定，直到实例化该函数为止，此时SFINAE可能会跳入以影响调用的重载设置.

I pulled this from STL's CppCon talk. TL;DW: when using the class's T type parameter, the compiler knows what that type is when instantiating the class template and would check for the type member at that point in time. With the extra TCopy type parameter being local to the function, the compiler cannot know for sure until the function is instantiated, at which point SFINAE can jump in to affect the overload set for the call.

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