Question

我正在尝试读取XML中的元素并将其存储在 struct，需要将此数组的指针传递给其他函数。但是我在gnu中编译时出现问题，错误消息：error：无法将'myRec'转换为uint32_t {aka unsigned int}'作为回报返回* recs;

Please help.

试图在不使用malloc的情况下设置myRec recs [count]，并得到无效指针的错误

struct myRec
{
std::string one;
std::string two;
std::string three;
std::string four;
std::string five;
std::string six;
};

uint32_t count = 0;

XMLDocument doc;
doc.LoadFile(pFilename);
XMLElement* parent = doc.FirstChildElement("a");
XMLElement* child = parent->FirstChildElement("b");
XMLElement* e = child->FirstChildElement("c");

for (e = child->FirstChildElement("c"); e; e = e->NextSiblingElement("c"))
{
    count++;
}

std::cout << "\n""Count = " << count << std::endl;

recs = (myRec *)malloc(6 *count * sizeof(myRec));

XMLElement *row = child->FirstChildElement();
if (count > 0)
{
    --count;
    count = (count < 0) ? 0 : count;
    for (uint32_t i = 0; i <= count; i++)
    {
        while (row != NULL)
        {
            std::string six;
            six = row->Attribute("ID");
            recs[i].six = six;

            XMLElement *col = row->FirstChildElement();
            while (col != NULL)
            {
                std::string sKey;
                std::string sVal;
                char *sTemp1 = (char *)col->Value();
                if (sTemp1 != NULL) {
                    sKey = static_cast<std::string>(sTemp1);
                }
                else {
                    sKey = "";
                }
                char *sTemp2 = (char *)col->GetText();
                if (sTemp2 != NULL) {
                    sVal = static_cast<std::string>(sTemp2);
                }
                else {
                    sVal = "";
                }
                if (sKey == "one") {
                    recs[i].one = sVal;
                }
                if (sKey == "two") {
                    recs[i].two = sVal;
                }
                if (sKey == "three") {
                    recs[i].three = sVal;
                }
                if (sKey == "four") {
                    recs[i].four = sVal;
                }
                if (sKey == "five") {
                    recs[i].five = sVal;
                }
                col = col->NextSiblingElement();
            }// end while col
            std::cout << "\n""one = " << recs[i].one << "\n"" two= " << recs[i].two << "\n""three = " << recs[i].three << "\n""four = " << recs[i].four << "\n""five = " << recs[i].five << "\n""six = " << recs[i].six << std::endl;
            row = row->NextSiblingElement();
        }// end while row
    }
}
else
{
    std::cout << "Failed to find value, please check XML! \n" << std::endl;
}
return *recs;

期望返回指向数组的指针

I declared it as 
std::string getxmlcontent(const char* pFilename);
myRec*recs= NULL;

function:
std::string readxml::getxmlcontent(const char* pFilename)
{
}

not sure if it is the right way as I am quite new to c++

Answer 1

您犯了一些错误，您可能应该得到good C++ book并进行一些学习

在C ++中，使用new

recs = new myRec[6*count];

代替

recs = (myRec *)malloc(6 *count * sizeof(myRec));

C ++程序中malloc的问题在于它不会调用构造函数，因此结构中包含的所有字符串均无效，并且（很可能）您的程序在运行时会崩溃。

我不清楚您为什么需要6*count，这似乎是因为您的结构中有六个字符串。如果是这样的话，那就太让人困惑了，真的只需要

recs = new myRec[count];

您将获得6*count字符串，因为这是您声明结构的方式。

sKey = sTemp1;

代替

sKey = static_cast<std::string>(sTemp1);

不需要强制转换，将char*分配给std::string是完全合法的。

最后，如果您想返回一个指针，则只需返回指针

return recs;

不是

return *recs;

但是，您尚未在发布的代码中包含函数签名。我怀疑还有另一个错误，但是除非您发布如何声明此函数，否则我无法告诉您。

使用tinyXML2读取XML时如何返回指向结构数组的指针

1 个答案: