C++设置空指针？

Question

我如何将其设置为空；

LookupTable<Product *> table;

我尝试的一切都说

1 IntelliSense:不存在合适的构造函数来将“int”转换为“LookupTable”

查找表是这样的:

#ifndef LOOKUPTABLE_H
#define LOOKUPTABLE_H

#include <iostream>
#include <string>

using namespace std;

#define MAXRANGE 10

template <class T>
class LookupTable
{
private:
    T *aptr[MAXRANGE];
    int rangeStart[MAXRANGE];
    int rangeEnd[MAXRANGE];
    int numRanges;

public:
    T defaultValue;
    bool failedRangeCheck;
    std::string failReason;


    // Constructor
    LookupTable() 
    {   
        numRanges = 0; 
        defaultValue = T();
    }      

    void addRange(int start, int end)
    {
        std::cout << "Created a new range...  Start: " << start << " / End: " << end << endl;
        failedRangeCheck = false;

        //lines omitted because not working anyway

        if ( !failedRangeCheck )
        {
            //set ranges
            rangeStart[numRanges] = start;
            rangeEnd[numRanges] = end;

            //build new generic array with end-start+1 positions
            //set pointer to point to it
            aptr[numRanges] = new T[ end - start + 1 ];
            numRanges++;
        }
        else
        {
            std::cout << "Range overlapped another range." << endl;
            std::cout << failReason << endl;
        }
    }

    T &operator[](int value)     // Overloaded [] operator
    {
        for ( int i = 0; i < numRanges; i++ )
        {
            if ( (value >= rangeStart[i]) && (value <= rangeEnd[i]) )
            {
                return aptr[i][value - rangeStart[i]];
            }
        }

        return defaultValue;
    }

    ~LookupTable()
    {
         delete[] aptr;
         numRanges = 0;     
    }

};
#endif

Answer 1

在您的示例中，table不是指针。它是一个 LookupTable 类型的对象，其中 LookupTable 是一个模板类。在这种情况下，您恰好正在制作一个专门用于 Product * 的 LookupTable。 s，但如果你这样做，你会得到同样的错误 LookupTable< int > table = 0;