Re: overloading of ","
Michael DOUBEZ wrote:
josh a ?crit :
Hi, I coded the following but It does not return what I expect, why?
#include <iostream>
using namespace std;
class Other
{
public:
int i;
Other(int x=1)
{
i = x;
}
Other *operator-> () { return this;}
Other &operator+ (Other t)
{
i += t.i;
return *this;
}
Other &operator,(Other oth)
{
i = oth.i;
return *this;
}
};
int main()
{
Other o0, o1, o2(4), o3(5);
o0->i = 100;
cout << o0.i << "\n" << o0->i << "\n";
// HERE it returns 5 AND not 6 WHY ???????????????????
Other ox = (o1 + o1, o3 = o2 + o1);
Because you have 5.
The expression evaluates:
Other ox = ( (o1 + o1) , (o3 = o2 + o1) );
Or with names
Other ox = o1.operator+(o1).operator,(o3.operator=(o2.operator+(o1)));
Since o3 is 5, then o1 is also 5 and ox is 5.
The reason is overloaded operator, doesn't have the same precedence as
POD operator,. Is is very confusing.
That is not the only thing confusing! Your use of operator,() is the so
obfuscated! Why on Earth would you want to do something like that?
What possible reason could you have?
On a side note, why isn't operator,() the same precedence as the regular
',' operator? I thought that all user defined operators had the same
precedence as the builtins.
Adrian
--
==========================================================
Adrian Hawryluk BSc. Computer Science
----------------------------------------------------------
Specialising in: OOD Methodologies in UML
OOP Methodologies in C, C++ and more
RT Embedded Programming
__--------------------------------------------------__
----- [blog: http://adrians-musings.blogspot.com/] -----
'--------------------------------------------------------'
My newsgroup writings are licensed under the Creative
Commons Attribution-Noncommercial-Share Alike 3.0 License
http://creativecommons.org/licenses/by-nc-sa/3.0/
==========================================================