In the c++ primer ,i get a program.
A class's name is TT,and it define the operator overload!
TT first; //constructor
TT second(30);//constructor
TT thrid(40://constructor
first=second.op erator+;
the question is the fourth line is all right?
maybe that is :
first=second.op erator+(third);
which one is right? 6 2084
jay wrote: In the c++ primer ,i get a program. A class's name is TT,and it define the operator overload!
TT first; //constructor TT second(30);//constructor TT thrid(40://constructor first=second.op erator+;
the question is the fourth line is all right?
maybe that is : first=second.op erator+(third);
which one is right?
i think u r right:)
jay wrote: In the c++ primer ,i get a program. A class's name is TT,and it define the operator overload!
TT first; //constructor TT second(30);//constructor TT thrid(40://constructor first=second.op erator+;
the question is the fourth line is all right?
Even the third line is not alright. The fourth line is definitely missing
some parentheses.
maybe that is : first=second.op erator+(third);
which one is right?
Depending on how the operator is defined, either could be right.
V
--
Please remove capital As from my address when replying by mail
< TT first; //constructor
< TT second(30);//constructor
< TT thrid(40://constructor
< first=second.op erator+;
< the question is the fourth line is all right?
< maybe that is :
< first=second.op erator+(third);
< which one is right?
For example:
class TT
{
public:
TT() : Value( 0 ) {}
TT( int value ) : Value( value ) {}
TT operator+() // first overloading
{
return TT( this->Value );
}
TT operator+ ( const TT& op ) // second overloading
{
return TT( this->Value + op.Value );
}
protected:
int Value;
};
//----------------------------------------------------------------------------------
int _tmain(int argc, _TCHAR* argv[])
{
TT first; //constructor
TT second(30); //constructor
TT third(40); //constructor
first = second.operator +(); // first overloading (it
maybe should be operator =)
first = second + third; // second overloading
std::cout << first.Value << std::endl;
return 0;
}
This is perfect legally code!
So, if at your 4-th statement, add some paranthesis, the code looks
fine
It's only depending on how your operator+ is defined!
asterisc <ra*******@fian system.com> wrote: int _tmain(int argc, _TCHAR* argv[]) { TT first; //constructor TT second(30); //constructor TT third(40); //constructor
first = second.operator +(); // first overloading (it maybe should be operator =) first = second + third; // second overloading std::cout << first.Value << std::endl;
return 0; }
This is perfect legally code!
It's not because _TCHAR is not defined anywhere. Also are you
missing a main function. _tmain is non-standard.
regards
--
jb
(reply address in rot13, unscramble first)
>> This is perfect legally code! It's not because _TCHAR is not defined anywhere. Also are you missing a main function. _tmain is non-standard.
Sorry, i use to work on windows platform, and i made a... quick
project.
It automaticaly includes:
#include <iostream>
#include <tchar.h>
But, ignore that, and replace: _tmain with main, and _TCHAR with char
;)
Anyways, the main focus was the overloaded operators, and their
callings, not.. the main entry
"jay" <gu************ @gmail.com> wrote in message
news:11******** **************@ 38g2000cwa.goog legroups.com... In the c++ primer ,i get a program. A class's name is TT,and it define the operator overload!
TT first; //constructor TT second(30);//constructor TT thrid(40://constructor first=second.op erator+;
the question is the fourth line is all right?
You can find a way to make it work, if you correct the third line to TT
third(40); and write a program like the following:
#include <iostream>
class TT
{
private:
void (*m_p)(const TT&);
public:
TT()
: m_p(NULL)
{}
TT(int n)
: m_p(NULL)
{}
static void operator+(const TT& rhs) // dubious and contrived
{
std::cout << "Blah\n";
}
TT& operator=(void (*p)(const TT&))
{
m_p = p;
return *this;
}
void f()
{
(*m_p)(*this);
}
};
int main()
{
TT first;
TT second(30);
TT third(40);
first = second.operator +;
first.f();
return 0;
}
Is it a good idea? Almost invariably not (never say "never", of course!) But
it does go to show that it can be done if you really want to :)
HTH,
Stu
maybe that is : first=second.op erator+(third);
which one is right? This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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