Question

我有一个constexpr数组，如下所示：

constexpr std::array<int, 4> BASES_TO_CHECK = { 8, 16, 32, 64 };

我想做些类似的事情：

std::array<std::thread, BASES_TO_CHECK.size()> calc;
for(size_t i = 0; i<BASES_TO_CHECK.size(); ++i)
{
    calc[i]=std::thread(calculate<BASES_TO_CHECK[i]>,  std::ref(recordMap[BASES_TO_CHECK[i]]),  std::ref(counterMap.at(BASES_TO_CHECK[i])),  std::ref(done));
}

但是，由于该变量用作模板参数，因此无法使用。我最终这样做了：

std::array<std::thread, BASES_TO_CHECK.size()> calc = {
    std::thread(calculate<BASES_TO_CHECK[0]>,  std::ref(recordMap[BASES_TO_CHECK[0]]),  std::ref(counterMap.at(BASES_TO_CHECK[0])),  std::ref(done)),
    std::thread(calculate<BASES_TO_CHECK[1]>,  std::ref(recordMap[BASES_TO_CHECK[1]]),  std::ref(counterMap.at(BASES_TO_CHECK[1])),  std::ref(done)),
    std::thread(calculate<BASES_TO_CHECK[2]>,  std::ref(recordMap[BASES_TO_CHECK[2]]),  std::ref(counterMap.at(BASES_TO_CHECK[2])),  std::ref(done)),
    std::thread(calculate<BASES_TO_CHECK[3]>,  std::ref(recordMap[BASES_TO_CHECK[3]]),  std::ref(counterMap.at(BASES_TO_CHECK[3])),  std::ref(done))
};

这可行，但要依靠我不更改BASES_TO_CHECK中元素的数量，而无需手动更新初始化calc数组的代码部分。

Answer 1

template<std::size_t... i>
std::array<std::thread, BASES_TO_CHECK.size()> gen_impl(std::index_sequence<i...>) {
    return {
                std::thread(calculate<BASES_TO_CHECK[i]>,
                            std::ref(recordMap[BASES_TO_CHECK[i]]),
                            std::ref(counterMap.at(BASES_TO_CHECK[i])),
                            std::ref(done)
                           )...
           };
}

auto calc = gen_impl(std::make_index_sequence<BASES_TO_CHECK.size()>{});

如何以静态方式使用另一个constexpr数组初始化一个数组

1 个答案: