QVariant conversions (.toString() .toInt()) performance.

[jsulm]

@Mike-Fidel You could write a simple benchmark to measure the difference.
I think var.toString() will be slower as it creates a new QString instance each time it is called.

[Mike Fidel]

@jsulm This is the benchmark I did:

int main(int argc, char *argv[])
{
    QElapsedTimer variantTimer;
    QElapsedTimer stringTimer;

    QVariant var = QVariant("My String");

    QString string = "My String";



    QString stringTest;
    QString varTest;


    stringTimer.start();
    for (int i = 0 ; i < 100000 ; i++)
    {
          stringTest = string;
    }

    std::cout << "string  took " << stringTimer.nsecsElapsed() << " nano seconds " << std::endl;



    variantTimer.start();
    for (int i = 0 ; i < 100000 ; i++)
    {
        varTest = var.toString();
    }

    std::cout << "Variant took " << variantTimer.nsecsElapsed() << " nano seconds " << std::endl;


}

Output I got:

string took 2084675 nano seconds
Variant took 4343839 nano seconds

However if you include the deceleration of QString inside the loop like so:

int main(int argc, char *argv[])
{
    QElapsedTimer variantTimer;
    QElapsedTimer stringTimer;

    QVariant var = QVariant("My String");

    QString string = "My String";



    //QString stringTest;
    //QString varTest;


    stringTimer.start();
    for (int i = 0 ; i < 100000 ; i++)
    {
          QString stringTest = string;
    }

    std::cout << "string  took " << stringTimer.nsecsElapsed() << " nano seconds " << std::endl;



    variantTimer.start();
    for (int i = 0 ; i < 100000 ; i++)
    {
        QString varTest = var.toString();
    }

    std::cout << "Variant took " << variantTimer.nsecsElapsed() << " nano seconds " << std::endl;


}

I get the following output:

string took 4607827 nano seconds
Variant took 2628095 nano seconds

This confuses me. It seems that Variant is quicker if you declare the variable inside the loop.

Also the conclusion I draw from this is that performance impact of the conversion methods are largely negligible. And if your implementation requires that you use an definition with the .toString() conversion(Such as QString x = myVar.toString()), you are better off using QVariant.

[jsulm]

@Mike-Fidel Probably I was wrong: QVariant already contains a QString

[J.Hilk]

Isn't QString and QVariant each Implicitly Shared ?

that means

for (int i = 0 ; i < 100000 ; i++)
    {
          stringTest = string;
    }

does basicaly nothing? Decrese and Increase of shared pointer count

where as

 for (int i = 0 ; i < 100000 ; i++)
    {
        varTest = var.toString();
    }

is an actual convertion => Construction of a QString and copy of data(from Qvariant)?

[Mike Fidel]

@jsulm said in QVariant conversions (.toString() .toInt()) performance.:

@Mike-Fidel Probably I was wrong: QVariant already contains a QString

So QVariant might be faster, but costs more memory?

[jsulm]

@Mike-Fidel It should not cost more memory because of implicit sharing as @J-Hilk mentioned.

[Mike Fidel]

@jsulm said in QVariant conversions (.toString() .toInt()) performance.:

@Mike-Fidel It should not cost more memory because of implicit sharing as @J-Hilk mentioned.

Could you elaborate? As far as I understand "Implicit Sharing" means that on assignment a pointer is passed, and only on write a copy occurs.

So in

for (int i = 0 ; i < 100000 ; i++)
    {
          stringTest = string;
    }

And in

 for (int i = 0 ; i < 100000 ; i++)
    {
        varTest = var.toString();
    }

The memory fingerprint of 'varTest' and 'stringTest' should be the same. But you said that QVariant already contains a QString, shouldn't it take up more memory?

Also How does implicit sharing explains QVariant being faster when the deceleration is made inside the loop?

[jsulm]

@Mike-Fidel QString uses implicit sharing as well. So, assigning a string to another one does not increase memory usage significantly.
Is it

QString varTest;
for (int i = 0 ; i < 100000 ; i++)
{
    varTest = var.toString();
}

or

for (int i = 0 ; i < 100000 ; i++)
{
    QString varTest = var.toString();
}

?

[Mike Fidel]

@jsulm

QString uses implicit sharing as well. So, assigning a string to another one does not increase memory usage significantly.

Yes I understand, However I'm confused by what you said that QVariant already contains QString. Because if it does, shouldn't it take up more memory than just QString ? Because QVaraint is QString and something else ?

[JonB]

@Mike-Fidel
@jsulm did use the word "significantly". I don't know whether someone is saying the whole of a QVariant is shared or just its content, but if it's the latter, yes, a QVaraint is a QString and something else, but the "something else" is relatively small, at least compared to, say, an "average" 42-byte string.

[Mike Fidel]

@JonB Got it, Thanks!