Are string.Equals() and == operator really same? [duplicate]

Answer 1

C# has two "equals" concepts: Equals and ReferenceEquals. For most classes you will encounter, the == operator uses one or the other (or both), and generally only tests for ReferenceEquals when handling reference types (but the string Class is an instance where C# already knows how to test for value equality).

• Equals compares values. (Even though two separate int variables don't exist in the same spot in memory, they can still contain the same value.)
• ReferenceEquals compares the reference and returns whether the operands point to the same object in memory.

Example Code:

var s1 = new StringBuilder("str");
var s2 = new StringBuilder("str");
StringBuilder sNull = null;

s1.Equals(s2); // True
object.ReferenceEquals(s1, s2); // False
s1 == s2 // True - it calls Equals within operator overload
s1 == sNull // False
object.ReferenceEquals(s1, sNull); // False
s1.Equals(sNull); // Nono!  Explode (Exception)

Answer 2

In addition to Jon Skeet's answer, I'd like to explain why most of the time when using == you actually get the answer true on different string instances with the same value:

string a = "Hell";
string b = "Hello";
a = a + "o";
Console.WriteLine(a == b);

As you can see, a and b must be different string instances, but because strings are immutable, the runtime uses so called string interning to let both a and b reference the same string in memory. The == operator for objects checks reference, and since both a and b reference the same instance, the result is true. When you change either one of them, a new string instance is created, which is why string interning is possible.

By the way, Jon Skeet's answer is not complete. Indeed, x == y is false but that is only because he is comparing objects and objects compare by reference. If you'd write (string)x == (string)y, it will return true again. So strings have their ==-operator overloaded, which calls String.Equals underneath.

Answer 3

There are plenty of descriptive answers here so I'm not going to repeat what has already been said. What I would like to add is the following code demonstrating all the permutations I can think of. The code is quite long due to the number of combinations. Feel free to drop it into MSTest and see the output for yourself (the output is included at the bottom).

This evidence supports Jon Skeet's answer.

Code:

[TestMethod]
public void StringEqualsMethodVsOperator()
{
    string s1 = new StringBuilder("string").ToString();
    string s2 = new StringBuilder("string").ToString();

    Debug.WriteLine("string a = \"string\";");
    Debug.WriteLine("string b = \"string\";");

    TryAllStringComparisons(s1, s2);

    s1 = null;
    s2 = null;

    Debug.WriteLine(string.Join(string.Empty, Enumerable.Repeat("-", 20)));
    Debug.WriteLine(string.Empty);
    Debug.WriteLine("string a = null;");
    Debug.WriteLine("string b = null;");

    TryAllStringComparisons(s1, s2);
}
private void TryAllStringComparisons(string s1, string s2)
{
    Debug.WriteLine(string.Empty);
    Debug.WriteLine("-- string.Equals --");
    Debug.WriteLine(string.Empty);
    Try((a, b) => string.Equals(a, b), s1, s2);
    Try((a, b) => string.Equals((object)a, b), s1, s2);
    Try((a, b) => string.Equals(a, (object)b), s1, s2);
    Try((a, b) => string.Equals((object)a, (object)b), s1, s2);

    Debug.WriteLine(string.Empty);
    Debug.WriteLine("-- object.Equals --");
    Debug.WriteLine(string.Empty);
    Try((a, b) => object.Equals(a, b), s1, s2);
    Try((a, b) => object.Equals((object)a, b), s1, s2);
    Try((a, b) => object.Equals(a, (object)b), s1, s2);
    Try((a, b) => object.Equals((object)a, (object)b), s1, s2);

    Debug.WriteLine(string.Empty);
    Debug.WriteLine("-- a.Equals(b) --");
    Debug.WriteLine(string.Empty);
    Try((a, b) => a.Equals(b), s1, s2);
    Try((a, b) => a.Equals((object)b), s1, s2);
    Try((a, b) => ((object)a).Equals(b), s1, s2);
    Try((a, b) => ((object)a).Equals((object)b), s1, s2);

    Debug.WriteLine(string.Empty);
    Debug.WriteLine("-- a == b --");
    Debug.WriteLine(string.Empty);
    Try((a, b) => a == b, s1, s2);
#pragma warning disable 252
    Try((a, b) => (object)a == b, s1, s2);
#pragma warning restore 252
#pragma warning disable 253
    Try((a, b) => a == (object)b, s1, s2);
#pragma warning restore 253
    Try((a, b) => (object)a == (object)b, s1, s2);
}
public void Try<T1, T2, T3>(Expression<Func<T1, T2, T3>> tryFunc, T1 in1, T2 in2)
{
    T3 out1;

    Try(tryFunc, e => { }, in1, in2, out out1);
}
public bool Try<T1, T2, T3>(Expression<Func<T1, T2, T3>> tryFunc, Action<Exception> catchFunc, T1 in1, T2 in2, out T3 out1)
{
    bool success = true;
    out1 = default(T3);

    try
    {
        out1 = tryFunc.Compile()(in1, in2);
        Debug.WriteLine("{0}: {1}", tryFunc.Body.ToString(), out1);
    }
    catch (Exception ex)
    {
        Debug.WriteLine("{0}: {1} - {2}", tryFunc.Body.ToString(), ex.GetType().ToString(), ex.Message);
        success = false;
        catchFunc(ex);
    }

    return success;
}

Output:

string a = "string";
string b = "string";

-- string.Equals --

Equals(a, b): True
Equals(Convert(a), b): True
Equals(a, Convert(b)): True
Equals(Convert(a), Convert(b)): True

-- object.Equals --

Equals(a, b): True
Equals(Convert(a), b): True
Equals(a, Convert(b)): True
Equals(Convert(a), Convert(b)): True

-- a.Equals(b) --

a.Equals(b): True
a.Equals(Convert(b)): True
Convert(a).Equals(b): True
Convert(a).Equals(Convert(b)): True

-- a == b --

(a == b): True
(Convert(a) == b): False
(a == Convert(b)): False
(Convert(a) == Convert(b)): False
--------------------

string a = null;
string b = null;

-- string.Equals --

Equals(a, b): True
Equals(Convert(a), b): True
Equals(a, Convert(b)): True
Equals(Convert(a), Convert(b)): True

-- object.Equals --

Equals(a, b): True
Equals(Convert(a), b): True
Equals(a, Convert(b)): True
Equals(Convert(a), Convert(b)): True

-- a.Equals(b) --

a.Equals(b): System.NullReferenceException - Object reference not set to an instance of an object.
a.Equals(Convert(b)): System.NullReferenceException - Object reference not set to an instance of an object.
Convert(a).Equals(b): System.NullReferenceException - Object reference not set to an instance of an object.
Convert(a).Equals(Convert(b)): System.NullReferenceException - Object reference not set to an instance of an object.

-- a == b --

(a == b): True
(Convert(a) == b): True
(a == Convert(b)): True
(Convert(a) == Convert(b)): True

Answer 4

The Header property of the TreeViewItem is statically typed to be of type object.

Therefore the == yields false. You can reproduce this with the following simple snippet:

object s1 = "Hallo";

// don't use a string literal to avoid interning
string s2 = new string(new char[] { 'H', 'a', 'l', 'l', 'o' });

bool equals = s1 == s2;         // equals is false
equals = string.Equals(s1, s2); // equals is true

Answer 5

It is clear that tvi.header is not a String. The == is an operator that is overloaded by String class, which means it will be working only if compiler knows that both side of the operator are String.

Answer 6

An object is defined by an OBJECT_ID, which is unique. If A and B are objects and A == B is true, then they are the very same object, they have the same data and methods, but, this is also true:

A.OBJECT_ID == B.OBJECT_ID

if A.Equals(B) is true, that means that the two objects are in the same state, but this doesn't mean that A is the very same as B.

Strings are objects.

Note that the == and Equals operators are reflexive, simetric, tranzitive, so they are equivalentic relations (to use relational algebraic terms)

What this means: If A, B and C are objects, then:

(1) A == A is always true; A.Equals(A) is always true (reflexivity)

(2) if A == B then B == A; If A.Equals(B) then B.Equals(A) (simetry)

(3) if A == B and B == C, then A == C; if A.Equals(B) and B.Equals(C) then A.Equals(C) (tranzitivity)

Also, you can note that this is also true:

(A == B) => (A.Equals(B)), but the inverse is not true.

A B =>
0 0 1
0 1 1
1 0 0
1 1 1

Example of real life: Two Hamburgers of the same type have the same properties: they are objects of the Hamburger class, their properties are exactly the same, but they are different entities. If you buy these two Hamburgers and eat one, the other one won't be eaten. So, the difference between Equals and ==: You have hamburger1 and hamburger2. They are exactly in the same state (the same weight, the same temperature, the same taste), so hamburger1.Equals(hamburger2) is true. But hamburger1 == hamburger2 is false, because if the state of hamburger1 changes, the state of hamburger2 not necessarily change and vice versa.

If you and a friend get a Hamburger, which is yours and his in the same time, then you must decide to split the Hamburger into two parts, because you.getHamburger() == friend.getHamburger() is true and if this happens: friend.eatHamburger(), then your Hamburger will be eaten too.

I could write other nuances about Equals and ==, but I'm getting hungry, so I have to go.

Best regards, Lajos Arpad.