struct Baz {};
int bar() { return 0; }
float bar(int) { return 0.0f; }
void bar(Baz *) {}
void foo1(void (&)(Baz *)) {}
template <class T,class D>
auto foo2(D *d) -> void_t<decltype(d(std::declval<T*>()))> {}
int main() {
foo1(bar); // Works
foo2<Baz>(bar); // Fails
}
foo1没有问题,明确指定栏的类型.
但是,foo2通过SFINAE禁用其他版本的所有版本,无法使用以下消息进行编译：
main.cpp:19:5: fatal error: no matching function for call to 'foo2'
foo2<Baz>(bar); // Fails
^~~~~~~~~
main.cpp:15:6: note: candidate template ignored: couldn't infer template argument 'D'
auto foo2(D *d) -> void_t<decltype(d(std::declval<T*>()))> {}
^
1 error generated.
我的理解是,C不能同时解析重载函数的地址并执行模板参数推导.
是原因吗有没有办法使foo2< Baz>(bar); (或类似的东西)编译？
解决方法
WhenPis a function type,function pointer type,or pointer to member function type:
If the argument is an overload set containing one or more function templates,the parameter is treated as a non-deduced context.
If the argument is an overload set (not containing function templates),trial argument deduction is attempted using each of the members of the set. If deduction succeeds for only one of the overload set members,that member is used as the argument value for the deduction. If deduction succeeds for more than one member of the overload set the parameter is treated as a non-deduced context.
一旦扣除结束,SFINAE就会参与到游戏中,因此它无助于解决标准规则.
有关详细信息,可以在上面列出的项目符号的末尾看到示例.
关于你的最后一个问题：
Is there a way to make
foo2<Baz>(bar);(or something similar) compile ?
两种可能的选择：
foo2<Baz>(static_cast<void(*)(Baz *)>(bar));
这样你就可以从过载集中明确地选择一个函数.
>如果允许修改foo2,可以将其重写为：
template <class T,class R>
auto foo2(R(*d)(T*)) {}
它或多或少是你以前的,在这种情况下没有decltype和返回类型,你可以自由地忽略.其实你不需要使用任何SFINAE的功能去做,扣除就够了.在这种情况下,foo2< Baz>(bar);正确解决
