Problem with unique and a binary predicate?

[BCC]

What the heck am I doing wrong?

void MyClass::JoinIdenticalFingerprints(CFPrintList& list)
{
CFPrintList unique_list = list;
unique(unique_list.begin(), unique_list.end(), UniquePredicate);
}

bool MyClass::UniquePredicate(const CFPrint& x, const CFPrint& y)
{
return x == y;
}

I get an error:
C:\Program Files\Microsoft Visual Studio .NET
2003\Vc7\include\algorithm(547): error C2064: term does not evaluate to
a function taking 2 arguments

The example I followed was the help file from MSDN, looks pretty much
the same to me.

It must be something really simple... is there anything obviously wrong
with they way I create and call unique with a predicate??

Thanks

 

[Victor Bazarov]

What the heck am I doing wrong?

void MyClass::JoinIdenticalFingerprints(CFPrintList& list)
{
CFPrintList unique_list = list;
unique(unique_list.begin(), unique_list.end(), UniquePredicate);
}

bool MyClass::UniquePredicate(const CFPrint& x, const CFPrint& y)
{
return x == y;
}

I get an error:
C:\Program Files\Microsoft Visual Studio .NET
2003\Vc7\include\algorithm(547): error C2064: term does not evaluate to
a function taking 2 arguments

The example I followed was the help file from MSDN, looks pretty much
the same to me.

It must be something really simple... is there anything obviously wrong
with they way I create and call unique with a predicate??

Your "predicate" is probably a non-static member, which requires one
additional argument, "this", to perform its actions. Since the function
does not seem to do anything 'MyClass'-related, you could declare it
static (if you didn't). If my guess is not right, post more code (the
'MyClass' definition is one important piece missing here.

Victor

 

[BCC]

Your "predicate" is probably a non-static member, which requires one
additional argument, "this", to perform its actions. Since the function
does not seem to do anything 'MyClass'-related, you could declare it
static (if you didn't). If my guess is not right, post more code (the
'MyClass' definition is one important piece missing here.

Victor

Good guess!

Declaring it static worked.

Thanks!
B