Converting std::string to QString in constructor - c++

Question

I cannot understand why std::string converted into QString while passing it to constructor. Here is small example:
class StringHandler
{
private:
QString str;
public:
StringHandler(QString s): str(s) {}
};
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
std::string str = "string";
QString qstr(str); // it gives error there are no constructor QString(std::string)
StringHandler handler(QString(str));//it does not give an error. Why?
return a.exec();
}
EDIT:
class StringHandler
{
public:
StringHandler(QString s): str(s) {}
QString str;
};
int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
std::string str = "string";
StringHandler handler(QString(str));//Here should be error, but there no error. Why?
qDebug()<<handler.str; //Error is here: request for member 'str' in 'handler', which is of non-class type 'StringHandler(QString)'
return a.exec();
}

Answer 1

Say hello to the most vexing parse.
StringHandler handler(QString(str)); declares a function named handler that takes a QString and returns a StringHandler. Yes. Thanks to the C++ parsing rules.
Now the error message request for member 'str' in 'handler', which is of non-class type 'StringHandler(QString)' makes sense: handler is being treated like a function of type StringHandler(QString) and you're trying to access a member named str in it, but of course functions have no members so the compilation fails.
You can fix this by using the uniform initialization syntax:
StringHandler handler{QString(str)};
The above can't be parsed as a function declaration, and so the compilation should fail for the expected reason: no matching constructor QString(std::string).
For more info, see C++'s most vexing parse again.