Hi All,
I don't understand very well this part of MSDN:
"Derived classes that override GetHashCode must also override Equals to
guarantee that two objects considered equal have the same hash code;
otherwise, Hashtable might not work correctly."
Does any one know, why we must also override Equals,
Please give my an example:)
Thanks,
Stoyan 5 9554
Hi Stoyan,
The Hashtable type requires that if two objects are equal, they must have
the same hash code value : when a key/value pair is added to a Hashtable,
the Hashtable uses the hash code of the key to determine the bucket that
stores the key. The bucket is then searched for a key which Equals the
given key (there might be more than one key in the bucket). If you wonder
why Equals is needed to find the key in the bucket : hashes can collide
(i.e. two different objects might generate the same hash - a hash
collision). http://samples.gotdotnet.com/quickst...hashtable.aspx
Regards,
Bart
-- http://www.xenopz.com
"Stoyan" <St****@discussions.microsoft.com> wrote in message
news:FF**********************************@microsof t.com... Hi All, I don't understand very well this part of MSDN: "Derived classes that override GetHashCode must also override Equals to guarantee that two objects considered equal have the same hash code; otherwise, Hashtable might not work correctly." Does any one know, why we must also override Equals, Please give my an example:)
Thanks, Stoyan
Hi Bart,
I understand very well, if I override Equals, I must override and
GetHashCode() method.
but I don't understand for example:
I have two classes:
public class SomeType
{
public override int GetHashCode()
{ return 1;}
}
public class AnotherType
{
public override int GetHashCode()
{ return 1;}
}
SomeType a = new SomeType();
AnotherType b = new AnotherType();
Hashtable tbl = new Hashtable();
tbl.Add(a,a);
tbl.Add(b,b);
object a = tbl[a]; // this object is correct (SomeType)
object b = tbl[b]; // this object is correct (AnotherType)
In my case, why must also override Equals()?
I know, if hashcodes are different, Hashtable working better(faster)
Regards,
Stoyan
"Bart De Boeck" wrote: Hi Stoyan,
The Hashtable type requires that if two objects are equal, they must have the same hash code value : when a key/value pair is added to a Hashtable, the Hashtable uses the hash code of the key to determine the bucket that stores the key. The bucket is then searched for a key which Equals the given key (there might be more than one key in the bucket). If you wonder why Equals is needed to find the key in the bucket : hashes can collide (i.e. two different objects might generate the same hash - a hash collision). http://samples.gotdotnet.com/quickst...hashtable.aspx
Regards, Bart
-- http://www.xenopz.com "Stoyan" <St****@discussions.microsoft.com> wrote in message news:FF**********************************@microsof t.com... Hi All, I don't understand very well this part of MSDN: "Derived classes that override GetHashCode must also override Equals to guarantee that two objects considered equal have the same hash code; otherwise, Hashtable might not work correctly." Does any one know, why we must also override Equals, Please give my an example:)
Thanks, Stoyan
Stoyan wrote: Hi All, I don't understand very well this part of MSDN: "Derived classes that override GetHashCode must also override Equals to guarantee that two objects considered equal have the same hash code; otherwise, Hashtable might not work correctly."
For hashtables to operate, the following must be true:
Equals(x,y) -> x.GetHashCode() == y.GetHashCode()
x.GetHashCode() != y.GetHashCode() -> !Equals(x,y)
This effectivly joins the partitioning that Equals imposes on objects
into larger sets, which rather neatly are numbered..., ready to put in
an index'ed array. Cunning thing, hashing :)
The above makes a constraint on how you can re-define GetHashCode
without redefining Equals.
Stricly speaking, if you use (keep the default) reference-equality the
only constraint on GetHashCode is that it is consistent:
x == y -> x.GetHashCode(x) == y.GetHashCode(y).
Does any one know, why we must also override Equals,
Well, strictly speaking... it's not required, but...
Please give my an example:)
In short, Hashtables usually does linear comparison, using Equals, of
the lookup key with every entry with the same hash-value modulo the size
of the hashtable. This also shows why choosing a hash-function with good
distribution is very important.
For hashing to be *usefull*, one usually calculates the hashcode on the
basis of the *value* of an object, not the reference, and *that's* why
you need to override Equals.
A minimalistic example:
class Foo {
public readonly int X;
public Foo(int x) { this.X = x; }
public override int GetHashCode() { return X; }
}
If you were to insert Foo's into a hashtable:
IDictionary d = new Hashtable();
Foo foo = new Foo(0);
d.Add(foo, 0);
You would (probably) expect :
Foo foo2 = new Foo(0);
object x = d[foo];
object y = d[foo2];
to both return a boxed 0, but they *dont* foo.GetHashCode() == 0 and
foo2.GetHashCode() == 0, but !Equals(foo, foo2).
So you need to redefine Equals:
public override bool Equals(object other) {
return ((Foo)other).X == X;
}
To do comparison by value, instead of reference.
--
Helge
Thank you Helge :))
Best Regards,
Stoyan
"Helge Jensen" wrote: Stoyan wrote: Hi All, I don't understand very well this part of MSDN: "Derived classes that override GetHashCode must also override Equals to guarantee that two objects considered equal have the same hash code; otherwise, Hashtable might not work correctly."
For hashtables to operate, the following must be true:
Equals(x,y) -> x.GetHashCode() == y.GetHashCode() x.GetHashCode() != y.GetHashCode() -> !Equals(x,y)
This effectivly joins the partitioning that Equals imposes on objects into larger sets, which rather neatly are numbered..., ready to put in an index'ed array. Cunning thing, hashing :)
The above makes a constraint on how you can re-define GetHashCode without redefining Equals.
Stricly speaking, if you use (keep the default) reference-equality the only constraint on GetHashCode is that it is consistent:
x == y -> x.GetHashCode(x) == y.GetHashCode(y).
Does any one know, why we must also override Equals,
Well, strictly speaking... it's not required, but...
Please give my an example:)
In short, Hashtables usually does linear comparison, using Equals, of the lookup key with every entry with the same hash-value modulo the size of the hashtable. This also shows why choosing a hash-function with good distribution is very important.
For hashing to be *usefull*, one usually calculates the hashcode on the basis of the *value* of an object, not the reference, and *that's* why you need to override Equals.
A minimalistic example:
class Foo { public readonly int X; public Foo(int x) { this.X = x; } public override int GetHashCode() { return X; } }
If you were to insert Foo's into a hashtable:
IDictionary d = new Hashtable(); Foo foo = new Foo(0); d.Add(foo, 0);
You would (probably) expect :
Foo foo2 = new Foo(0); object x = d[foo]; object y = d[foo2];
to both return a boxed 0, but they *dont* foo.GetHashCode() == 0 and foo2.GetHashCode() == 0, but !Equals(foo, foo2).
So you need to redefine Equals:
public override bool Equals(object other) { return ((Foo)other).X == X; }
To do comparison by value, instead of reference.
-- Helge
as far as I know, one of the reasons that you should override Equals is that
might require that different objects (different references of the same type)
can be equal
so, based on your example :
public class SomeType
{
public override int GetHashCode()
{ return 1;}
int _field = 3;
}
if two instances of SomeType are equal (cause _field equals), you should
override Equals
-- http://www.xenopz.com
"Stoyan" <St****@discussions.microsoft.com> wrote in message
news:5B**********************************@microsof t.com... Hi Bart, I understand very well, if I override Equals, I must override and GetHashCode() method. but I don't understand for example: I have two classes: public class SomeType { public override int GetHashCode() { return 1;} } public class AnotherType { public override int GetHashCode() { return 1;} }
SomeType a = new SomeType(); AnotherType b = new AnotherType(); Hashtable tbl = new Hashtable(); tbl.Add(a,a); tbl.Add(b,b);
object a = tbl[a]; // this object is correct (SomeType) object b = tbl[b]; // this object is correct (AnotherType)
In my case, why must also override Equals()? I know, if hashcodes are different, Hashtable working better(faster)
Regards, Stoyan
"Bart De Boeck" wrote:
Hi Stoyan,
The Hashtable type requires that if two objects are equal, they must have the same hash code value : when a key/value pair is added to a Hashtable, the Hashtable uses the hash code of the key to determine the bucket that stores the key. The bucket is then searched for a key which Equals the given key (there might be more than one key in the bucket). If you wonder why Equals is needed to find the key in the bucket : hashes can collide (i.e. two different objects might generate the same hash - a hash collision). http://samples.gotdotnet.com/quickst...hashtable.aspx
Regards, Bart
-- http://www.xenopz.com "Stoyan" <St****@discussions.microsoft.com> wrote in message news:FF**********************************@microsof t.com... > Hi All, > I don't understand very well this part of MSDN: > "Derived classes that override GetHashCode must also override Equals to > guarantee that two objects considered equal have the same hash code; > otherwise, Hashtable might not work correctly." > Does any one know, why we must also override Equals, > Please give my an example:) > > Thanks, > Stoyan This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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