QList generic join() function with template

问题

I am trying to make a generic join() function for QList (like join() for QStringList) in order to make a toString() function for a QList of any type. This function takes a QList, a separator and a function to dertermine how to print items. Consider this code :

#include <QList>
#include <QDebug>

template <class T>
static QString join(const QList<T> &list, const QString &separator, const std::function< QString (const T &item) > toStringFunction)
{
    QString out;
    for(int i = 0; i<list.size(); i++)
        out+= (i ? separator : "") + toStringFunction(list[i]);
    return out;
}

int main(int argc, char *argv[])
{
    QList <double> list;
    list<<1.<<2.<<3.<<4.;
    int precision = 1;
    QString out = join(list, ",",[precision](const double &item)->QString{
                    return QString::number(item,'f',precision);
                    });

    qDebug()<<out;
    return 1;
}

Here the errors I have :

src\main.cpp(18): error C2672: 'join': no matching overloaded function found
src\main.cpp(20): error C2784: 'QString join(const QList<T> &,const QString &,const std::function<QString(const T &)>)': could not deduce template argument for 'const std::function<QString(const T &)>' from 'main::<lambda_f1fd4bbd6b8532d33a84751b7c214924>'
src\main.cpp(5): note: see declaration of 'join'

Clearly I dont care about this function, plenty of solutions to do it. But I don't understand what I am doing wrong with templates here.

could not deduce template argument ???

NB :

out = join<double>(list, ",",[precision](const double &item)->QString{
    return QString::number(item,'f',precision);
});

=> Works fine

const std::function<QString(const double &item)> toStringFunction = [precision](const double &item)->QString{
    return QString::number(item,'f',precision);
};
out = join(list, ",",toStringFunction);

=> Works fine

回答1:

I'm not sure what's going on with the C++ internals, but it does work with this declaration:

template <class T> 
static QString join(const QList<T> &list, 
                    const QString &separator, 
                    const std::function< QString (const typename QList<T>::value_type &) > toStringFunction)

I think QList can determine the template type from the list being passed, while the join template itself can't.

来源：https://stackoverflow.com/questions/57890027/qlist-generic-join-function-with-template

标签

templates

function-pointers

qlist