小编Til*_*tel的帖子

因为我是 Haskell 的新手，所以我对一切都不太熟悉。我尝试使用模式匹配来实现一个函数，但这不起作用，因为它给出了以下异常：Prelude.head 空列表。这是代码示例：

swapListToTowers::[[Bool]]->[[Bool]]
swapListToTowers [] = []
swapListToTowers xs = [head x| x <-xs]:swapListToTowers (cutOfFirstElements xs)


cutOfFirstElements::[[Bool]]->[[Bool]]
cutOfFirstElements [] = []
cutOfFirstElements xs = [tail x | x <- xs]

我认为捕获空列表的情况不太好，是吗？在此先感谢您的帮助！

假设我从 github API 调用到文件有以下 Base64 编码的字符串：

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我有一个问题,我无法在任何地方找到答案.但是,我必须展示一些代码:

  #include "Vector2D"

    class something
    {
        Vector2D * p_Position;
      public:
        something(){p_Position = new Vector2D;}
        ~something(){delete p_Position;}
    };

    int main()
{
    std::list<something> Somethinglist;

    Somethinglist.push_back(something());

    Somethinglist.clear();
    return 0;
}

因此,当涉及到.clear()函数时,这将导致断言失败.所以我试了几件事.首先,如果我不把它delete p_Position放在解构器中,这个代码就完全可以了.这是为什么？STL列表.clear()功能是否会自动销毁动态指针？或者一个相当直接的问题:我该如何修复此代码？