指向对象类型检查的c ++指针 [英] c++ pointers to objects type checking

问题描述

如何验证指针所指向的对象有效

How do I verify that an object being pointed by a pointer is valid

相关代码

LookupTable<Product *> table;
Product temp = *table[selection];
// if *table[selection] is not a product, program crashes...

这里是查找表:

#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

推荐答案

基于您在

Based on the code of LookupTable that you posted in a subsequent question, this question can finally be answered. (Really, I think you ought to just put that code here and remove the other question.)

table[selection]或者返回对表中某个条目的引用(如果找到了选择)，否则返回对默认初始化对象的引用.当LookupTable专用于指针类型(例如代码中的Product*)时，默认初始化的对象将是NULL指针.

table[selection] either returns a reference to an entry in the table (if the selection is found) or otherwise a reference to a default-initialized object. When LookupTable is specialized for a pointer type (such as the Product* in your code) then the default-initialized object will be a NULL pointer.

因此，对于代码中的LookupTable<Product*>，表达式table[selection]要么将结果指向表中找到的Product指针，要么为NULL Product指针.

So, for the LookupTable<Product*> in your code, the expression table[selection] is either going the result in a pointer to a Product found within the table or else a NULL Product pointer.

因此，您应该立即获取指针值并进行检查，而不是立即取消引用table[selection]的结果并将其分配给Product对象的尝试.

Consequently, instead of immediately dereferencing the result of table[selection] and trying to assign it to a Product object, you should actually take the pointer value and examine it.

这将通过类似于以下代码的代码来实现:

This would be accomplished with code similar to:

Product* result = table[selection];
if(result != NULL)
{
    Product temp = *result;
    // do something with temp, etc, etc
}
else
{
    cout << "invalid product code" << endl;
}

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