struct and passing array parameters

[Digitale_GFacchini]

@Digitale_GFacchini said in struct and passing array parameters:

dichiarative

Sorry sierdzio, but dichiarative it's a class. the class is a object. or no?

[SGaist]

Hi,

No, a class is a definition used to create objects AKA "instances".

[Digitale_GFacchini]

@sierdzio said in struct and passing array parameters:

dichiarative myObject;
myObject.settingDichiarative(argv);

ok so, i i want use a classe a need a istance.
Now i have a problem in parameter why? I have declared :

char* argv

into in file header and in file cpp where is the class dichiarative and his method

    dichiarative myObject;
    myObject.settingDichiarative(argv);

main.cpp:12:38: error: no matching function for call to 'dichiarative::settingDichiarative(char**&)'
myObject.settingDichiarative(argv);
^

[Digitale_GFacchini]

Ok hai have put:

myObject.settingDichiarative(*argv);

And so it's ok, but now i have error in method settingDichiarative.

C:\Generale\OneDrive - ELICAT SRL - 03978700288\Developer\QtApplicazioni\youtube_and_web\EVILEG\SQLiteDbAcccesListModel/dichiarative.cpp:14: undefined reference to `dichiarative::paramingresso::arg_db_name'

This error for every idem of structure declared into .h

struct paramingresso {
        static QString arg_db_type;
        static QString arg_db_driver;
        static QString arg_db_name;
        static QString arg_db_path;
    };

and the method is :

void dichiarative::settingDichiarative(char* argomenti)
{
    dichiarative::paramingresso::arg_db_type = QStringLiteral("%1").arg(argomenti[1]);
    dichiarative::paramingresso::arg_db_driver = QStringLiteral("%1").arg(argomenti[2]);
    dichiarative::paramingresso::arg_db_name = QStringLiteral("%1").arg(argomenti[3]);
    dichiarative::paramingresso::arg_db_path = QStringLiteral("%1").arg(argomenti[4]);
}

In this monts i have sty but it's too much arguments and every time I try to apply I block myself for errors .. Be patient then when I understand bele the logic I will be able to reciprocate. At least I hope

[sierdzio]

This is - again - because you have no object / variable of type "paramingresso". In your "dichiarative" class, add this:

public:
  paramingresso myStruct;

And now modify your method like so:

void dichiarative::settingDichiarative(char* argomenti)
{
    myStruct.arg_db_type = QStringLiteral("%1").arg(argomenti[1]);
    myStruct.arg_db_driver = QStringLiteral("%1").arg(argomenti[2]);
    myStruct.arg_db_name = QStringLiteral("%1").arg(argomenti[3]);
    myStruct.arg_db_path = QStringLiteral("%1").arg(argomenti[4]);
}

Also, as a side note - it's much easier to extract command line arguments using Qt methods. Check out https://doc.qt.io/qt-5/qcoreapplication.html#arguments or (a bit more advanced) https://doc.qt.io/qt-5/qcommandlineparser.html#details

[Digitale_GFacchini]

Thanks for reference to extract command line. however I continue as an exercise with reading argv and setting up a structure. Are you telling me to instantiate the structure in the header file? If I understand correctly the result is:

undefined reference to `DichiarativeGlobali::paramingresso::arg_db_type'
undefined reference to `DichiarativeGlobali::paramingresso::arg_db_driver'
undefined reference to `DichiarativeGlobali::paramingresso::arg_db_name'
undefined reference to `DichiarativeGlobali::paramingresso::arg_db_path'

P:S. I have changed aold name class in DichiarativeGlobali.

[VRonin]

@Digitale_GFacchini said in struct and passing array parameters:

Ok hai have put:
myObject.settingDichiarative(*argv);

This is wrong, you should change settingDichiarative(char* argomenti) to settingDichiarative(char** argomenti) and call myObject.settingDichiarative(argv);.

[Digitale_GFacchini]

Thanks VRonin but .. . Why? in main.cpp i have :

int main(int argc, char *argv[])

What difference from * to ** ??

[jsulm]

@Digitale_GFacchini said in struct and passing array parameters:

char *argv[]

this is same as "char **argv".
You should read a book or tutorial about C++ basics.

[Digitale_GFacchini]

thanks VRonin,
topic of * is "pointers". But why two ** ??
Do you know any good books in Italian? I'm reading everything in English and I was not using it for years.

[SGaist]

char * = array of char
char ** = array of array of char

[Digitale_GFacchini]

Thank SGaist