c++:将向量转换为元组
如何将 std::vector 转换为 std::tuple ?我有
How can I convert std::vector to std::tuple ? I have
class T { };
int cnt = 3;
vector<T*> tv;
for (int i = 0; i < cnt; ++i) {
tv.push_back(new T());
}
我想得到
auto tp = std::tie(*tv[0], *tv[1], *tv[2]);
我怎样才能得到这个 tp ?如果cnt足够大，我不能手动写这个tp.
How can I get this tp ? If cnt is big enough, I can't write this tp manually.
std::vector<
ConvConnection<
decltype(inputLayer),
decltype(*C1[0]),
decltype(*Conn1Opt[0]),
RandomInitialization<arma::mat>,
arma::mat
>* > Conn1(6);
for (size_t i = 0; i < 6; ++i) {
Conn1.push_back(new ConvConnection<
decltype(inputLayer),
decltype(*C1[0]),
decltype(*Conn1Opt[0]),
RandomInitialization<arma::mat>,
arma::mat
>(inputLayer, *C1[i], *Conn1Opt[i], 5, 5));
}
这是代码.这里只有 6 个，但我还需要一些大小超过 100 的向量.我需要将此向量转换为元组.
This is the code. Here is just 6, but I also need some vector whose size is over 100. I need to convert this vector to a tuple.
推荐答案
一般情况下，您不能将 vector 转换为 tuple.但是，如果您要做的只是为某些 制作元组 是一个常量表达式，那么这可以通过索引序列技巧来实现:
Generally, you cannot convert a vector to a tuple. However, if all you're trying to do is make the tuple <f(0), f(1), ..., f(N-1)> for some N that is a constant-expression, then that is doable with the index sequence trick:
template <typename F, size_t... Is>
auto gen_tuple_impl(F func, std::index_sequence<Is...> ) {
return std::make_tuple(func(Is)...);
}
template <size_t N, typename F>
auto gen_tuple(F func) {
return gen_tuple_impl(func, std::make_index_sequence<N>{} );
}
我们可以这样使用:
// make a tuple of the first 10 squares: 0, 1, 4, ..., 81
auto squares = gen_tuple<10>([](size_t i){ return i*i;});
对于您的特定用例，这将是:
For your specific use-case, that would be:
auto connections = gen_tuple<6>([&](size_t i) {
return new ConvConnection<
decltype(inputLayer),
decltype(*C1[0]),
decltype(*Conn1Opt[0]),
RandomInitialization<arma::mat>,
arma::mat
>(inputLayer, *C1[i], *Conn1Opt[i], 5, 5);
});
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