QString url = prepareFeedURL(itemType, cType); means



  • Please tell me the meaning aboutt this code....

    QString url = prepareFeedURL(itemType, cType);
    parser->fetch(url);



  • [quote author="Prajnaranjan Das" date="1293517881"]Please tell me the meaning aboutt this code....

    QString url = prepareFeedURL(itemType, cType);
    parser->fetch(url);[/quote]

    How do you think that by just giving two lines of code someone can tell you what it does. It seems prepareFeedURL is a user defined function and what it does is not known.

    By the function names it seems the function gets feed url for a particular item type and then its stored in to the string url and sent to the parser to fetch it.

    You should not be asking such questions in a forum. It seems you are not willing to put any effort by your own.



  • I think you should try clicking F1 on functions on Qt Creator, each function is documented very well on Qt Assistant and this Qt Creator feature is perfect.



  • [quote author="Prajnaranjan Das" date="1293517881"]Please tell me the meaning aboutt this code....

    QString url = prepareFeedURL(itemType, cType);
    parser->fetch(url);[/quote]

    prepareFeedURL seems to be a function that takes two arguments (of unknown type, since the prototype is missing) and returns a QString or a structure that can be auto-converted to a QString (e.g. a const char * or a QByteArray).

    parser seems to be an object that has a method fetch that takes one argument which is a QString, a reference to a QString or a const reference to a QString (or any other type, reference to other type or const reference to other type to which a QString can be auto-converted).

    Since neither the "Flying Spaghetti Monster":http://en.wikipedia.org/wiki/Flying_Spaghetti_Monster nor the "Pythia of Delphi":http://en.wikipedia.org/wiki/Pythia is available at the moment, we cannot tell you more.



  • thank you Volker and Milot Shala for your valuable reply....



  • [quote author="Prajnaranjan Das" date="1293544340"]thank you Volker and Milot Shala for your valuable reply....
    [/quote]

    You are welcome dude :)


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