c++ - 错误 : expression cannot be used as an function

Question

我是 lambda 的新手，我制作了自己的带有自定义比较器函数的二进制堆类。一切顺利，直到出现编译错误，我不知道如何修复。

我试着稍微改变一下我的代码行，而不是

this(capacity, [](int a, int b){return a - b;});

我改成了这样:

function<int(int, int)> cmp = [](int a, int b){return a - b;};
this(capacity, cmp);

我得到了相同的结果。如何处理这个错误？

二叉堆类:

class binaryheap
{
private:
    int *heap;
    int size;
    int capacity;
    function<int(int, int)> cmp;
    int parent(int i);
    int left_child(int i);
    int right_child(int i);
    void swap(int *a, int *b);
    void heapify(int i);
public:
    binaryheap(int capacity);
    binaryheap(int capacity, const function<int(int, int)>& cmp);
    ~binaryheap();
    bool empty();
    int heap_size() const;
    int get_root() const;
    int extract_root();
    void decrease_key(int i, int value);
    void insert_key(int key);
    void delete_key(int i);
};

我的代码有编译错误的部分

binaryheap::binaryheap(int capacity)
{
    this(capacity, [](int a, int b){return a - b;});//binaryheap.cpp:51:58: error: expression cannot be used as a function
}

binaryheap::binaryheap(int capacity, const function<int(int, int)>& cmp)
{
    this->capacity = capacity;
    this->heap = new int[capacity + 1];
    this->size = 0;
    this->cmp = cmp;
}

Answer 1

我想你想使用委托(delegate)构造函数功能；所以

 binaryheap::binaryheap (int capacity)
     : binaryheap{capacity, [](int a, int b){return a - b;}}
  { }

或者，按照 melpomene 的建议(谢谢)，您可以删除此构造函数并为另一个构造函数中的第二个参数。