QJsonObject: Insert in a deep field

[fem_dev]

I would like to insert/append a QJsonObject inside a deep QJsonObject/Array field.
Example:

QJsonObject root;

QJsonArray my_array;

QJsonObject first;
first.insert("first", 1);

my_array.append(first);

root.insert("my_array", my_array);

Until here...all ok.

Now, I would like to append a second QJsonObject inside the my_array field.
Like:

QJsonObject second;
second.insert("second", 2);

root["my_array"].append(second); // Not working!

Why? How can I fix it?

[Gojir4]

@fem_dev Hi,

The issue here is that the compiler cannot know you are accessing a QJSonArray. QJsonValueRef QJsonObject::operator[](QStringView key) returns a QJsonValue, and not a QJsonArray, you can access the array using QJsonValue::toArray() const but then it returns a const array so you cannot modify it "on the fly"

If you have still access to your array instance:

my_array.append(first);
...
my_array.append(second);
root.insert("my_array", my_array);

But if you don't, you need to get the array, append your data, and then put it back in the QJsonObject.

QJsonArray myArray = root["my_array"].toArray();
myArray.append(second);
root["my_array"] = myArray;

[Gojir4]

@fem_dev Hi,

The issue here is that the compiler cannot know you are accessing a QJSonArray. QJsonValueRef QJsonObject::operator[](QStringView key) returns a QJsonValue, and not a QJsonArray, you can access the array using QJsonValue::toArray() const but then it returns a const array so you cannot modify it "on the fly"

If you have still access to your array instance:

my_array.append(first);
...
my_array.append(second);
root.insert("my_array", my_array);

But if you don't, you need to get the array, append your data, and then put it back in the QJsonObject.

QJsonArray myArray = root["my_array"].toArray();
myArray.append(second);
root["my_array"] = myArray;

[fem_dev]

@Gojir4 thank you!

If I use this:

QJsonArray myArray = root["my_array"].toArray(); // Object copying?
myArray.append(second);
root["my_array"] = myArray;

Question1:
It is copying the array? Or just the pointer value?
Is there a way to avoid object copying?

Question 2:
Ok, I understood that I need the toArray() method.
But now, why I can't do that?

 root["my_array"].toArray().append(second); // Not working too

[mrjj]

Hi
try
root["my_array"] = root["my_array"].toArray().append(second);

I think
root["my_array"].toArray() is a copy and not a reference/pointer so you are modifying a temp object and not the
original one.

[fem_dev]

@mrjj I tried:

root["my_array"] = root["my_array"].toArray().append(second);

But I got this compilation time error:

app.cpp:392:26: error: no viable overloaded '='
qjsonvalue.h:181:20: note: candidate function not viable: cannot convert argument of incomplete type 'void' to 'const QJsonValue' for 1st argument
qjsonvalue.h:182:20: note: candidate function not viable: cannot convert argument of incomplete type 'void' to 'const QJsonValueRef' for 1st argument

Question:
Is there another way to append() avoiding the object copying below? Using a pointer or a reference?

QJsonArray myArray = root["my_array"].toArray(); // Object copying?
myArray.append(second);
root["my_array"] = myArray;

[mrjj]

@fem_dev said in QJsonObject: Insert in a deep field:

QJsonValueRef

Ok it complains about incomplete type
so i wonder if it misses an include?

Anyway, @Gojir4 showed the working way.

• Is there another way to append() avoiding the object copying below? Using a pointer or a reference?
  Well it seems its a copy
  https://doc.qt.io/qt-5/qjsonvalue.html#toArray
  So unless it can return a ref than i think not.

[fcarney]

It seems like the JSON objects in Qt are designed to be assembled rather than modified:
https://doc.qt.io/qt-5/qtcore-serialization-savegame-example.html

Is there a word for this type of code design?

[mrjj]

@fcarney
Well its a value-orientated design utilizing the implicit sharing Qt provides making the copying cheap.
However, it does allow ref for some of the constructs via QJsonValueRef

Im not sure it has a concrete name other than maybe
Surprising :)