[英]Deducing return and parameter type from std::function passed as a template function argument?
I've been looking around SO for a while now but cannot find quite the answer that I am looking for - this question is probably the closest to what I am thinking. 我一直在四处寻找SO,但是找不到我一直在寻找的答案- 这个问题可能最接近我的想法。
In a sentence: is it possible to declare a template function that takes a parameter of an std::function
and deduce template parameters for both the return type and the parameter types of the function? 一句话:是否可以声明一个采用
std::function
参数的模板函数,并为该std::function
的返回类型和参数类型推导出模板参数? Example: 例:
//this works to pass the std::function in
template<class T>
void doSomething(std::function<T> f) {
f();
}
//this is more what i am looking for - can R and P be deduced automatically - does not work!
template<class R, class P>
void doSomethingElse(std::function<R(P)> f) {
f();
}
Is this because the function signature or function type is considered one thing in itself, and as such cannot be "broken" up? 这是因为函数签名或函数类型本身被视为一件事情,因此不能被“分解”吗? I realise there are
decltype
and std::result_of
but cannot think how I might use them here. 我意识到这里有
decltype
和std::result_of
但是无法思考我在这里如何使用它们。
As an additional point, how might I extend the second example to have multiple parameters and deduction, using variadic templates? 另外,如何使用可变参数模板将第二个示例扩展为具有多个参数和推导?
template<class R, class P>
void doSomethingElse(std::function<R(P)> f) {
f(P{});
}
Will work, but it only works if you pass a std::function
to the function and that function has one non void parameter. 可以,但是仅当您将
std::function
传递给该函数并且该函数具有一个非void参数时,该函数才有效。 This is kind of limiting though. 不过,这是一种限制。 You can use
您可以使用
template<class R, class... Args, class... Ts>
void doSomethingElse(std::function<R(Args...)> f, Ts&&... args) {
f(std::forward<Args>(args)...);
}
Which will take any std::function
and the arguments for it and calls them as if you did it in the call site. 它将接受任何
std::function
及其参数,并像在调用站点中一样调用它们。 This is still limiting though because the call site requires you use a std::function
so you can't pass it anything implicitly convertible to a std::function
. 但是这仍然是有限制的,因为调用站点要求您使用
std::function
所以您不能将任何隐式可转换为std::function
传递给它。
With C++17 and class template argument deduction (CTAD) this is no longer an issue though. 使用C ++ 17和类模板参数推论 (CTAD)时,这不再是问题。 We can create an overload that takes any type, and then construct a std::function using CTAD to fill in the types for us.
我们可以创建任何类型的重载,然后使用CTAD构造一个std :: function来为我们填充类型。 That would look like
看起来像
template<class Func, class... Args>
void doSomethingElse(Func&& f, Args&&... args) {
doSomethingElse(std::function{std::forward<Func>(f)}, std::forward<Args>(args)...);
}
template<class R, class... Args, class... Ts>
void doSomethingElse(std::function<R(Args...)> f, Ts&&... args) {
f(std::forward<Args>(args)...);
}
And now anything that isn't a std::function
will go to void doSomethingElse(Func&& f, Args&&... args)
, get converted to a std::function
, and get passed to void doSomethingElse(std::function<R(Args...)> f, Args&&... args)
so you can use the return type and argument(s) type(s) in there. 现在,所有不是
std::function
都将变为void doSomethingElse(Func&& f, Args&&... args)
,被转换为std::function
,并传递给void doSomethingElse(std::function<R(Args...)> f, Args&&... args)
因此您可以在其中使用返回类型和参数类型。
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