英文:
Specify function interface in template
问题
我有一个模板类，用于转发函数：
template <class T>class Foo{template <typename Func, typename... Args>void Bar(Func function, Args&&... args) {T t;(t->*function)(std::forward<Args>(args)...);}}
这个代码是可以工作的，但是有没有办法检查 Func 是否包含在 T 的接口中，即 T::Func？现在，如果我输入一个错误的函数，编译器会报错说 t->*function 不存在，但是我想在函数接口中指定它应该是一个 T::Func。然而，我似乎无法找到正确的语法。
英文:
I have a template class that forwards a function:
template <class T>class Foo{template <typename Func, typename... Args>void Bar(Func function, Args&&... args) {T t;(t->*function)(std::forward<Args>(args)...);}}
This works, but is there any way to check whether Func is contained in the interface of T, so T::Func? Now, if I enter a wrong function the compiler would complain that t->*function doesn't exist, but I would like to specify in the function interface that it should be a T::Func. However I can't seem to get the syntax for this right.
答案1
得分: 3
这不是真的必要在签名中指定这一点，因为它无论如何都会在错误的情况下编译失败，但你可以像这样指定成员函数指针：
template <class T>class Foo{public:template <typename Func, typename... Args>void Bar(Func T::*function, Args&&... args) {T t;(t.*function)(std::forward<Args>(args)...);}};
调用它的方式如下：
struct S{void func(int) {}};void freeFunc(int) {}int main(){Foo<S> f;f.Bar(&S::func, 42); //OK//f.Bar(&freeFunc, 42); //ERROR}
英文:
It's not really necessary to specify this in the signature, since it will fail to compile for the wrong cases anyway, but you could specify member function pointers like this:
template <class T>class Foo{public:template <typename Func, typename... Args>void Bar(Func T::*function, Args&&... args) {T t;(t.*function)(std::forward<Args>(args)...);}};
And to call it:
struct S{void func(int) {}};void freeFunc(int) {}int main(){Foo<S> f;f.Bar(&S::func, 42); //OK//f.Bar(&freeFunc, 42); //ERROR}
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