# Object.ReferenceEquals(Object, Object)Object.ReferenceEquals(Object, Object)Object.ReferenceEquals(Object, Object)Object.ReferenceEquals(Object, Object) Method

## Definição

Determina se as instâncias de Object especificadas são a mesma instância.Determines whether the specified Object instances are the same instance.

public:
static bool ReferenceEquals(System::Object ^ objA, System::Object ^ objB);
public static bool ReferenceEquals (object objA, object objB);
static member ReferenceEquals : obj * obj -> bool
Public Shared Function ReferenceEquals (objA As Object, objB As Object) As Boolean

#### Parâmetros

objA
Object Object Object Object

O primeiro objeto a ser comparado.The first object to compare.

objB
Object Object Object Object

O segundo objeto a ser comparado.The second object to compare.

#### Retornos

true se objA for a mesma instância que objB ou se ambos forem null; caso contrário, false.true if objA is the same instance as objB or if both are null; otherwise, false.

## Exemplos

O exemplo a seguir ReferenceEquals usa para determinar se dois objetos são da mesma instância.The following example uses ReferenceEquals to determine if two objects are the same instance.

using namespace System;
int main()
{
Object^ o = nullptr;
Object^ p = nullptr;
Object^ q = gcnew Object;
Console::WriteLine( Object::ReferenceEquals( o, p ) );
p = q;
Console::WriteLine( Object::ReferenceEquals( p, q ) );
Console::WriteLine( Object::ReferenceEquals( o, p ) );
}

/*

This code produces the following output.

True
True
False

*/
object o = null;
object p = null;
object q = new Object();

Console.WriteLine(Object.ReferenceEquals(o, p));
p = q;
Console.WriteLine(Object.ReferenceEquals(p, q));
Console.WriteLine(Object.ReferenceEquals(o, p));

// This code produces the following output:
//   True
//   True
//   False
Public Class App
Public Shared Sub Main()
Dim o As Object = Nothing
Dim p As Object = Nothing
Dim q As New Object
Console.WriteLine(Object.ReferenceEquals(o, p))
p = q
Console.WriteLine(Object.ReferenceEquals(p, q))
Console.WriteLine(Object.ReferenceEquals(o, p))
End Sub
End Class
' This code produces the following output:
'
' True
' True
' False
'

## Comentários

Ao contrário Equals do método e do operador de igualdade ReferenceEquals , o método não pode ser substituído.Unlike the Equals method and the equality operator, the ReferenceEquals method cannot be overridden. Por isso, se você quiser testar duas referências de objeto para igualdade e não tiver certeza sobre a implementação do Equals método, poderá chamar o ReferenceEquals método.Because of this, if you want to test two object references for equality and you are unsure about the implementation of the Equals method, you can call the ReferenceEquals method.

No entanto, o valor de ReferenceEquals retorno do método pode parecer anormal nestes dois cenários:However, the return value of the ReferenceEquals method may appear to be anomalous in these two scenarios:

• Ao comparar tipos de valor.When comparing value types. Se objA ReferenceEquals e objB forem tipos de valor, eles serão incluídos antes de serem passados para o método.If objA and objB are value types, they are boxed before they are passed to the ReferenceEquals method. Isso significa que objA se e objB representar a mesma instância de um tipo de valor, o ReferenceEquals método, no falseentanto, retornará, como mostra o exemplo a seguir.This means that if both objA and objB represent the same instance of a value type, the ReferenceEquals method nevertheless returns false, as the following example shows.

int int1 = 3;
Console.WriteLine(Object.ReferenceEquals(int1, int1));
Console.WriteLine(int1.GetType().IsValueType);

// The example displays the following output:
//       False
//       True

Public Module Example
Public Sub Main
Dim int1 As Integer = 3
Console.WriteLine(Object.ReferenceEquals(int1, int1))
Console.WriteLine(int1.GetType().IsValueType)
End Sub
End Module
' The example displays the following output:
'       False
'       True

Para obter informações sobre tipos de valores de boxing, consulte boxing e unboxing.For information on boxing value types, see Boxing and Unboxing.

• Ao comparar cadeias de caracteres.When comparing strings. Se objA ReferenceEquals true and objB forem cadeias de caracteres, o método retornará se a cadeia de caracteres for interna.If objA and objB are strings, the ReferenceEquals method returns true if the string is interned. Ele não executa um teste para igualdade de valor.It does not perform a test for value equality. No exemplo a seguir, s1 e s2 são iguais porque são duas instâncias de uma única cadeia de caracteres interna.In the following example, s1 and s2 are equal because they are two instances of a single interned string. No entanto s4 , s3 e não são iguais, porque embora tenham valores de cadeia de caracteres idênticos, essa cadeia de caracteres não é interna.However, s3 and s4 are not equal, because although they are have identical string values, that string is not interned.

String s1 = "String1";
String s2 = "String1";
Console.WriteLine("s1 = s2: {0}", Object.ReferenceEquals(s1, s2));
Console.WriteLine("{0} interned: {1}", s1,
String.IsNullOrEmpty(String.IsInterned(s1)) ? "No" : "Yes");

String suffix = "A";
String s3 = "String" + suffix;
String s4 = "String" + suffix;
Console.WriteLine("s3 = s4: {0}", Object.ReferenceEquals(s3, s4));
Console.WriteLine("{0} interned: {1}", s3,
String.IsNullOrEmpty(String.IsInterned(s3)) ? "No" : "Yes");

// The example displays the following output:
//       s1 = s2: True
//       String1 interned: Yes
//       s3 = s4: False
//       StringA interned: No

Module Example
Public Sub Main()
Dim s1 As String = "String1"
Dim s2 As String = "String1"
Console.WriteLine("s1 = s2: {0}", Object.ReferenceEquals(s1, s2))
Console.WriteLine("{0} interned: {1}", s1,
If(String.IsNullOrEmpty(String.IsInterned(s1)), "No", "Yes"))

Dim suffix As String = "A"
Dim s3 = "String" + suffix
Dim s4 = "String" + suffix
Console.WriteLine("s3 = s4: {0}", Object.ReferenceEquals(s3, s4))
Console.WriteLine("{0} interned: {1}", s3,
If(String.IsNullOrEmpty(String.IsInterned(s3)), "No", "Yes"))
End Sub
End Module
' The example displays the following output:
'       s1 = s2: True
'       String1 interned: Yes
'       s3 = s4: False
'       StringA interned: No