0

我正在尝试将字符串传递给类 mutator 并将私有成员设置为该字符串这里是发送字符串的代码

void parseTradePairs(Exchange::Currency *curr, std::string *response, int begin, int exit)
{
    int start;
    int end;
    string temp;
    string dataResponse;
    CURL *tempCurl;
    initializeCurl(tempCurl);
    int location = response->find("marketid", begin);
    if(location <= exit)
    {
        start = location + 11;
        begin = response->find("label", start);
        end = begin - start - 3;
        findStrings(start, end, temp, response);
        getMarketInfo(tempCurl, temp, dataResponse);
        curr->_coin->setExch(temp); // here is the line of code that is sending the string
        dataResponse >> *(curr->_coin);
        curr->_next = new Exchange::Currency(curr, curr->_position + 1);
        parseTradePairs(curr->_next, response, begin, exit);
    }

}

这是 coin 类中的 mutator，它接收字符串并将其分配给 _exch

void Coin::setExch(string exch)
{
    _exch = exch;
}

我已经通过它并确保 exch 中有字符串。“105”，但只要它击中 _exch = exch; 我得到了阅读违规。我也尝试作为指针传递。我不认为它应该超出范围。并且类中的字符串变量在默认构造函数中初始化为零，但除非我试图从中读取而不是写入它，否则这应该很重要。

/* defualt constructor */
Coin::Coin() 
{
    _id = "";
    _label = "";
    _code= "";
    _name = "";
    _marketCoin = "";
    _volume = 0;
    _last = 0;
    _exch = "";
}

Exchange::Exchange(std::string str)
{
    _exch = str;
    _currencies = new Currency;

    std::string pair;
    std::string response;
    CURL *curl;
    initializeCurl(curl);
    getTradePairs(curl, response);
    int exit = response.find_last_of("marketid");
    parseTradePairs(_currencies, &response, 0, exit);

}

int main(void)
{
    CURL *curl;
    string str;
    string id;
    Coin coin1;


    initializeCurl(curl);   

    Exchange ex("cryptsy");



    curl_easy_cleanup(curl);

    system("pause");

    return 0;
}
class Exchange
{
    public:

        typedef struct Currency
        {
            Currency(Coin *coin, Currency *next, Currency *prev, int position) : _coin(coin), _next(next), _prev(prev), _position(position) {}
            Currency(Currency *prev, int position) : _prev(prev), _position(position), _next(NULL), _coin(&Coin()){}
            Currency() : _next(NULL), _prev(NULL), _position(0) {}
            Coin *_coin;
            Currency *_next;
            Currency *_prev;
            int _position;
        };

        /* constructor and destructor */
        Exchange();
        Exchange(std::string str);
        ~Exchange();

        /* Assignment operator */
        Exchange& operator =(const Exchange& copyExchange);

        /* Parse Cryptsy Pairs */
        friend void parseTradePairs(Currency *curr, std::string *response, int begin, int exit);

    private:        
        std::string _exch;
        Currency *_currencies;
};

这是我将其更改为修复它的内容。typedef 结构货币

{
            Currency(Coin *coin, Currency *next, Currency *prev, int position) : _coin(coin), _next(next), _prev(prev), _position(position) {}
            Currency(Currency *prev, int position) : _prev(prev), _position(position), _next(NULL), _coin(&Coin()){}
            Currency()
            {
                _next = NULL;
                _prev = NULL;
                _position = 0;
                _coin = new Coin();
            }
            Coin *_coin;
            Currency *_next;
            Currency *_prev;
            int _position;
        };
4

1 回答 1

0

您的默认构造函数Currency是：

Currency() : _next(NULL), _prev(NULL), _position(0) {}

并且似乎没有为_coin. 您的Exchange::Exchange(std::string str)构造函数还可以：

_currencies = new Currency;

它调用 的默认构造函数Currency，然后传递给parseTradePairs()，所以很可能是这一行：

curr->_coin->setExch(temp);

因此试图取消引用一个未初始化的指针，结果你会遇到访问冲突。

于 2013-10-29T03:50:12.653 回答