xtensor - 如何用复杂的数据类型填充 xtensor 数组

Question

在这里转述，但这就是问题所在

xt::pyarray< std::complex<double> > output;
output = 0; // fails to compile
output = double(0); // also fails to compile
output = complex<double>(0); // succeeds

然而这很好

std::complex<double> foo;
foo = double(0);

问题是我正在编写一个接受任意 xtensor 数组的函数，例如。

template<typename xtarray_t>
void my_function(xtarray_t &output, const xtarray_t &input) {
    output = 0;
}

所以总的来说这个

output = complex<double>(0);

会失败。我看过这个，但不确定这是否已实现，或者如何使用它，找不到文档。但总的来说，我认为它仍然应该与 std::complex 一起使用。

谢谢你的帮助！

编译器输出：

xtensor-python/include/xtensor-python/pyarray.hpp:360:20: note: candidate function not viable: no known conversion from 'double' to 'const xt::pyarray<std::__1::complex<double>, xt::layout_type::dynamic>::self_type' (aka 'const pyarray<std::__1::complex<double>, (xt::layout_type)0>') for 1st argument 
    self_type& operator=(const self_type& rhs);

xtensor-python/include/xtensor-python/pyarray.hpp:363:20: note: candidate function not viable: no known conversion from 'double' to 'xt::pyarray<std::__1::complex<double>, xt::layout_type::dynamic>::self_type' (aka 'pyarray<std::__1::complex<double>, (xt::layout_type)0>') for 1st argument
    self_type& operator=(self_type&& e) = default;

xtensor-python/include/xtensor-python/pyarray.hpp:369:20: note: candidate template ignored: could not match 'xexpression<type-parameter-0-0>' against 'double'
    self_type& operator=(const xexpression<E>& e);

Answer 1

答案是这样做

using value_t = typename xtarray_t::value_type;
output = value_t(0);