'(' 标记之前的 C++ 错误预期主表达式

Question

我正在尝试创建一个程序，该程序从用户输入中获取 N 个随机节点，并创建一个随机整数，将其放入二叉树中，然后复制到优先级队列中。整数成为每个节点的键，另一个整数计算键的频率。当我复制到优先级队列时遇到问题，因为我得到重复项并且我需要删除它们。我试图通过节点构造函数创建一个集合，但在 .cpp 文件中出现上述错误。

#include <iostream>
#include <random>
#include <ctime>
#include <queue>
#include <set>
#include <functional>
#include <algorithm>
#include<list>
#include "Q7.h"

using namespace std;

int main()
{
node * root=NULL;
node z;
int n,v;

vector<int> first;
vector<int>::iterator fi;

default_random_engine gen(time(NULL));
cout<<"how many values? "; cin>>n;

for(int i=0; i<n; i++)
{  (v=gen()%n);
first.push_back(v);

    if(root==NULL){root = node(set(v));}///This is where I get the error!!
    else{
      root->addnode(v);
       }
}

z.unsortedRemoveDuplicates(first);

cout<<"Binary Tree in a depth first manner with Duplicates removed!"<<endl;
for ( fi = first.begin() ; fi != first.end(); ++fi{cout<<"Node "<<*fi<<endl;}
cout<<"-------------------"<<endl;
root->display();
cout<<"-------------------"<<endl;
cout<<"-------------------"<<endl;
 root->display_Queue1();
cout<<"-------------------"<<endl;

return 0;
 }





my .h file

class node
{
public:
node(){left=NULL; right=NULL; ct = 1;}
node set(int v) {val = v; left=NULL; right=NULL; ct=1;}
node (int Pri, int cat)
: val(Pri), ct(cat) {}

friend bool operator<(//sorts queue by lowest Priority
const  node& x, const  node& y)  {
    return x.val < y.val;
}

friend bool operator>(//sorts queue by greatest Priority
const node& x, const node& y) {
    return x.ct > y.ct;
}

friend ostream&//prints out queue later
operator<<(ostream& os, const node& Pri) {
return os  <<"my value = "<<Pri.val<<" occured "<<Pri.ct<<" times";
}

int unsortedRemoveDuplicates(vector<int>& numbers)
{
node set<int> seenNums; //log(n) existence check
auto itr = begin(numbers);
while(itr != end(numbers))
{
    if(seenNums.find(*itr) != end(seenNums)) //seen? erase it
        itr = numbers.erase(itr); //itr now points to next element
    else
    {
        seenNums.insert(*itr);
        itr++;
    }
}
return seenNums.size();
}
priority_queue<node, vector<node>, greater<node> > pq;
priority_queue<node, vector<node>, less<node> > pq1;

void addnode(int v)
{
   if(v==val){ct++;}
   pq.emplace(node (set (v)));///No error here for set with constructor why??
   pq.emplace(node (set (v)));
    if(v<val)
        {
        if(left==NULL){left=new node(set(v));
            }
        else{left->addnode(v);
            }
        }
    else
        {
        if(right==NULL){right = new node (set(v));
            }
        else{right->addnode(v);
            }
        }
}

int display()
    {
        if(left!=NULL){left->display();}

        cout<<"frequency  "<<ct<<" value"<<val<<endl;

        if(right!=NULL){right->display();}
    }

void display_Queue()
    {
        cout << "0. size: " << pq.size() << '\n';
        cout << "Popping out elements from Pqueue..."<<'\n';
        while (!pq.empty())
        {
        cout << pq.top() <<  endl;
        pq.pop();
        }
        cout << '\n';
    }
        void display_Queue1()
    {
        cout << "0. size: " << pq1.size() << '\n';
        cout << "Popping out elements from Pqueue..."<<'\n';
        while (!pq1.empty())
        {
        cout << pq1.top() <<  endl;
        pq1.pop();
        }
        cout << '\n';
    }

private:
int val;      ///value in that node
int ct;
///ct = count of that value
node * left;
node * right;
};

Answer 1

恭喜，这一行:

root = node(set(v));

你已经发现为什么这里的人经常说 avoid using using namespace std; .这被解释为:

root = static_cast<node>(std::set(v));

而不是你想要的，这可能是:

root = new node();
root->set(v);

首先，请注意，我们需要使用 new，因为我们正在创建一个新的 node，而不是尝试将 node 转换为 node，这也会给出另一个关于尝试为指针赋值的编译器错误。

接下来，请注意，您不会在头文件中收到错误，因为那里没有 using namespace std;，并且由于它在成员函数中，因此行:

void node::addnode(int v)
{
    //...
    pq.emplace(node (set (v)));///No error here for set with constructor why??
    //...
}

被解释为:

pq.emplace(static_cast<node>(this->set(v)));

但是，这是您真正想要做的吗？

---

此外，我会将构造函数更改为:

public:
    node (int Pri = 0, int cat = 1)
        : val(Pri), ct(cat), left(NULL), right(NULL) {}
    // DELETED node (int Pri, int cat)

因此你可以这样做:

root = new node(v);

它会像我认为的那样工作。