Question

我正在使用SFINAE机制,以推断出一种类型。 <>代码>Resolve<T>: 类型缩略语

class MyClass {};

template<typename>
struct void_ { typedef void check; };

template<typename T, typename = void>
struct Resolve { typedef T type; };

template<typename T>
struct Resolve <T, typename void_<typename T::yes>::check> {
    typedef MyClass type;
};

现在,我有一个简单的测试班,即:

struct B1 { typedef int yes; }; // 1
struct B2 { typedef int yes; }; // 2
struct D1 {};  // 3
struct D2 : B1 {};  // 4
struct D3 : B1, B2 {}; // 5 <----

根据以下逻辑,上述测试的结果应是:

Resove<B1>::type = MyClass
Resove<B2>::type = MyClass
Resove<D1>::type = D1
Resove<D2>::type = MyClass
Resove<D3>::type = MyClass or compiler error (due to ambiguity between B1, B2)

Strangely, in test case (5) it doesn t happen so. The result is,

Resolve<D3>::type = D3;

谁可以解释,什么是魔法>? 不能忘记汇编错误是一种符合标准的行为? http://ideone.com/xAzHV“rel=“nofollow”>demo。

Answer 1

Why would you expect a compiler error? You know SFINAE stands for Substitution Failure Is Not An Error right?

替换<代码>T时,该表述变得模糊不清。由于SFINAE,这一失败不被视为错误,而你作为候选人的专业被简单删除。它产生一个汇编错误。

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