QJsonObject: Notation error

fem_dev

Hi,

I'm using QJsonObject a lot, but now I got an notation error that is appearing only in one single Qt class of my project.

I'm always using this notation type:

double x = json["a"]["b"]["c"].toDouble();

So, inside my Qt class I typed:

double start = config["temperature"]["start"].toDouble();

But I got this error message:

error: no viable overloaded operator[] for type 'QJsonValueRef'

If I change the JSON notation to this line below, it works good:

double start = config["temperature"].toObject().value("start").toDouble();

Why this is happening? Why I can use the first notation above?

My class:

#include <QJsonObject>

class Step_Density
{
private:
    QJsonObject _config;

public:
    Step_Density();
};

SGaist

Hi,

What does temperature contain ?

fem_dev

@SGaist , the config is a QJsonObject.

The content of config is:

{
    "temperature": 
        {
            "start": 0.0,
            "end": 100.0
        }
}

What is wrong in the first (clean) notation?

mranger90

This is curious, but I think the problem is that operator[] can return either a QJsonValue or a QJsonValueRef.
This seems to work:

    qDebug() << "Header info is " << static_cast<QJsonValue>(configObject["Header"])["Tag"].toString();

So I think it's picking QJsonValueRef, since the documentation for QJsonValue clearly states that operator[] is
Equivalent to calling toObject().value(key).

fem_dev

@mranger90 thank you,

I type:

qDebug() << "Header info is " << static_cast<QJsonValue>(_config["Header"])["temperature"].toString();

and I got:

Header info is  ""

Yes, the operator [] is Equivalent to calling toObject().value(key), and I'm using this operator [] all the time in my project with QJsonObject.
My doubt is: Why in this Step_Density class the operator [] is returning a QJsonValueRef and not a QJsonValue as always was.

UPDATE 1:
Just to help to understand the error I remove the start node and I got no errors. Like:

double start = config["temperature"].toDouble();

So, I think that the problem is related with the second JSON node...
What it ca be?

mranger90

This works for me (Qt 5.13.0, ubuntu 19.04)


    QJsonObject configObject = jsonDoc.object();
    if (configObject.isEmpty())
    {
        qDebug() << "Json file is empty or root type is not an object (i.e. {})." << configObject;
    }

    qDebug() << "Header info is " << static_cast<QJsonValue>(configObject["Header"])["Temperature"].toDouble();

where the json file is:

{
    "Header": {
        "Tag": "Release 1.0",
        "Temperature": 42.24
    }
}

fem_dev

@mranger90, I understand your notation, but I got no point about why this is happening now?

Please, look my screen.
I uncommment the "wrong line" after the Debug starts only to be more visible in this image.

Could you help me to use the first notation?

PS.: the type of the private member class called _T.start is double

mranger90

The first notation does not work because, for reasons that are beyond my level of C++ expertise, the return of QJsonObject::operator[] is being interpreted as a QJsonValueRef, NOT a QJsonValue. Hence the cast I made.
The only other suggestion is to do it in 2 steps:

QJsonValue tempVal = _config["Temperature"];
_T.start = tempVal["start"].toDouble();
_T.end = tempVal["end"].toDouble();

fem_dev

@mranger90 said in QJsonObject: Notation error:

he only other suggestion is to do it in 2 steps:

Yes, I think that you are right...and your code works good. Thank you.

Well..I can use it...no problems! I just want to understand why this is hapenning...
I will let this post opened, ok?

Leonardo

I know this is an old topic, but I got here from Google, so this might be helpful for others in the future. If we take a look at the qjsonobject.h file, we see:

    QJsonValue operator[] (const QString &key) const;
    QJsonValue operator[] (QLatin1String key) const { return value(key); }
    QJsonValueRef operator[] (const QString &key);
    QJsonValueRef operator[] (QLatin1String key);

What determines the return type is the const keyword. This means that to use the syntax discussed in this thread, we need a const variable.