Error: expression cannot be used as an function

Question

我是 lambda 的新手，我用自定义比较器函数制作了自己的二进制堆 class。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);

我得到了同样的结果。如何处理这个错误？

二进制堆class:

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（谢谢）的建议，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添加默认值 ([](int a, int b){return a - b;})。

Error: expression cannot be used as an function

Error: expression cannot be used as an function

c++

c++11

lambda

std-function

delegating-constructor