join_rows()函数的正确类型是什么? [英] What's the right type for a join_rows() function?
问题描述
我写了一个函数,将两个2D的行连接起来数组:
template <typename S, typename T>
Array<typename S::Scalar, Dynamic, Dynamic> join_rows(const ArrayBase<S> & A, const ArrayBase<T> & B) {
Array<typename S::Scalar, Dynamic, Dynamic> C (A.rows(), A.cols()+B.cols());
C << A, B;
return C;
}
我想写一个更通用的函数,可以连接两个以上的数组
I would like to write a more general function that can join more than two arrays.
它应该能够使用任何可迭代的容器,例如。 std :: list
或 std :: vector
,所以我将使用模板模板参数计。
It should be able to work with any iterable container, eg. std::list
or std::vector
, so I would use a template template paratemeter.
我可以轻松地通过两个for循环来纠正函数体,这不是问题,我只是在努力弄清楚该函数的正确类型是什么。
I can easily right the function body with two for loops, that's not the issue, I'm just struggling to figure out what the right type for such a function would be.
(ps。我什至不确定我上面的代码是否具有最佳类型,但似乎能胜任工作)
(ps. I'm not even sure if my above code has the best type, but it seems to do the job)
推荐答案
我不确定如何声明任意 Array
s的向量,但是您可以实现将一个或更多参数直接传递给它。通常,这是通过递归调用自身,处理每个连续的参数来完成的:
I'm not sure how to declare a vector of arbitrary Array
s, but you can implement a function template that combines one or more arguments directly passed to it. This is typically done by calling itself recursively, processing each successive argument:
// end case (one argument): just forward the same array
template <typename T>
T&& join_rows(T&& A) {
return std::forward<T>(A);
}
// main function template: two or more arguments
template <typename S, typename T, typename... R>
Array<typename S::Scalar, Dynamic, Dynamic> join_rows(const ArrayBase<S>& A,
const ArrayBase<T>& B,
const ArrayBase<R>&... rest) {
Array<typename S::Scalar, Dynamic, Dynamic> C(A.rows(), A.cols()+B.cols());
C << A, B;
return join_rows(C, rest...); // call with the first two arguments combined
}
示例以说明用法:
int main() {
Array<int, 1, 3> arr1 = {1, 2, 3};
Array<int, 1, 2> arr2 = {4, 5};
Array<int, 1, 4> arr3 = {9, 8, 7, 6};
cout << join_rows(arr1, arr2, arr3.reverse()) << endl; // 1 2 3 4 5 6 7 8 9
return 0;
}
如果要限制一个参数,请 join_rows
仅接受 Eigen :: Array
s,然后添加 std :: enable_if
检查 ArrayBase< T>
基类:
If you want to restrict the one-argument join_rows
to only accept Eigen::Array
s, add an std::enable_if
checking for an ArrayBase<T>
base class:
template <typename T>
std::enable_if_t<std::is_base_of<ArrayBase<std::decay_t<T>>,std::decay_t<T>>::value, T&&>
join_rows(T&& A) {
return std::forward<T>(A);
}
对于大型 Array
s,可能会有更有效的方法来实现。您可能会返回一个仅分配一个新的 Array
对象的代理对象。
For large Array
s, there might be more efficient ways to implement this. You could probably return a proxy object that will only allocate one new Array
object.
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