Why does no operator< exist for two QStrings?
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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?
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Hi
I cant guess reason for the design choices but others have discussed it
http://stackoverflow.com/questions/16281014/operator-for-qstring0『Please use the Topic Tools button to mark as Solved
Upvote the answer(s) that helped you to solve the issue』1 Replies
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@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);
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@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-qstringI think you are sitting on a different boat...I was referring to operator< (less than)
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@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.
