Why does no operator< exist for two QStrings?

[Asperamanca]

Background: I get a compile error in the following code line:

return m_Id.operator< (other.m_Id);

Both m_Id and other.m_Id are of type QString.
Compiler output (MinGW) is:

error C2664: 'bool QString::operator <(const QByteArray &) const': cannot convert argument 1 from 'const QString' to 'QLatin1String'

That made me scratch my head. Looking up help, I realized that only the following operator< overloads are defined for QString:

bool operator<(QLatin1String other) const
bool operator<(const char *other) const
bool operator<(const QByteArray &other) const

...as well as this global function:

bool operator<(const char *s1, const QString &s2)

Why is that? And what's the best way to fix it?

[mrjj]

Hi
I cant guess reason for the design choices but others have discussed it
http://stackoverflow.com/questions/16281014/operator-for-qstring

[Taz742]

@Asperamanca
i dont understand.
you need to sort Arraylist or Vector?

bool comperator(QString a, QString b){
return a < b;
}

Qsort(array.begin(),array.end(),comperator);

[Asperamanca]

@mrjj said in Why does no operator< exist for two QStrings?:

Hi
I cant guess reason for the design choices but others have discussed it
http://stackoverflow.com/questions/16281014/operator-for-qstring

I think you are sitting on a different boat...I was referring to operator< (less than)

[Asperamanca]

@Taz742 said in Why does no operator< exist for two QStrings?:

@Asperamanca
i dont understand.
you need to sort Arraylist or Vector?

bool comperator(QString a, QString b){
return a < b;
}

Qsort(array.begin(),array.end(),comperator);

That gave me the hint. The code it not from me, and I didn't see the obvious: By writing

return m_Id.operator< (other.m_Id);

instead of

return m_Id < other.m_Id;

the code explicitly calls one of the overloads in QString, instead of any globally defined function.