set and operators

[Giulio Veronesi]

Hi all,

struct MyStruct {
int x, y;
};

bool operator<(const MyStruct& a1, const MyStruct& a2)
{
return ( (a1.x < a2.x) ||
( (a1.x == a2.x) && (a2.y < a2.y) ) );
}

typedef set<MyStruct> MySet;

Which other operators have I to overload to manage MySet? More
precisaly, when I insert something in the set, if the x field is equal
to another x field of another structure previous inserted in the set,
the second one is not inserted. I would like that a structure is not
inserted if and only if both its x and y filed are equals to x and y of
another structure in the set.

Thanks in advance,
Giulio

 

[Victor Bazarov]

struct MyStruct {
int x, y;
};

bool operator<(const MyStruct& a1, const MyStruct& a2)
{
return ( (a1.x < a2.x) ||
( (a1.x == a2.x) && (a2.y < a2.y) ) );
}

typedef set<MyStruct> MySet;

Which other operators have I to overload to manage MySet?

None. Assignment is useful, but the default one should suffice.

> More
precisaly, when I insert something in the set, if the x field is equal
to another x field of another structure previous inserted in the set,
the second one is not inserted. I would like that a structure is not
inserted if and only if both its x and y filed are equals to x and y of
another structure in the set.

All that should be expressed in the operator <. In your case, for a set
of your structs, two structures k and m are considered equivalent iff
(!(k < m) && !(m < k)) evaluates to true.

V

 

[Markus Moll]

Hi

bool operator<(const MyStruct& a1, const MyStruct& a2)
{
return ( (a1.x < a2.x) ||
( (a1.x == a2.x) && (a2.y < a2.y) ) );

Look carefully for an error in the above line...

}

Markus

 

[John Harrison]

Hi

Giulio Veronesi wrote:




Look carefully for an error in the above line...

Nope don't see it.

john

 

[=?iso-8859-1?Q?Ali_=C7ehreli?=]

Nope don't see it.

It is amazing how some bugs just don't want to reveal themselves. One of
the 2s should be 1...

Ali

 

[Markus Moll]

Nope don't see it.

Okay:

"a2.y < a2.y" is always false.
That makes "(a1.x == a2.x) && (a2.y < a2.y)" false.
Therefore, effectively, you return a1.x < a2.x

A simple typo, but it breaks your order...

hth
Markus

 

[John Harrison]

John Harrison wrote:




Okay:

"a2.y < a2.y" is always false.
That makes "(a1.x == a2.x) && (a2.y < a2.y)" false.
Therefore, effectively, you return a1.x < a2.x

A simple typo, but it breaks your order...

Aargh! I blame those spurious parens.

john