问题
Consider this code:
int a = 0;
short b = 0;
int c = 0;
object a1 = a;
object b1 = b;
object c1 = c;
Console.WriteLine(1);
//comparing primitives - int vs. short
Console.WriteLine(a == b);
Console.WriteLine(b == a);
Console.WriteLine(a.Equals(b));
Console.WriteLine(b.Equals(a));
Console.WriteLine(2);
//comparing objects - int vs. int
Console.WriteLine(c1 == a1);
Console.WriteLine(a1 == c1);
Console.WriteLine(c1.Equals(a1));
Console.WriteLine(a1.Equals(c1));
Console.WriteLine(3);
//comparing objects - int vs. short
Console.WriteLine(a1 == b1);
Console.WriteLine(b1 == a1);
Console.WriteLine(a1.Equals(b1)); //???
Console.WriteLine(b1.Equals(a1));
It prints this output:
1
True
True
True
False
2
False
False
True
True
3
False
False
False
False
What I know; what is clear
Section 2: ==
operator returns true when used with objects only if it compares one object in memory referenced by two different names (not very frequent, but might happen). Equals()
method compare content (value) of the objects. It is mentioned in many answers on this site.
Section 1: Using ==
operator, compiler converts ‘smaller’ type to ‘bigger’ (short
to int
) and compares primitive values. Order of operands (variables) doesn’t matter. Result of Equals()
in the last line might be confusing, because it returns false (doesn’t compare values), but it is understood. The order matter here. As learned in this answer, the best overload must be selected. It is selected by the type of the first variable: short.Equals(short)
. But then int
cannot be converted to ‘smaller’ type (short
), therefore there is no comparison made and method returns false.
Questions:
- Is my understanding above correct?
- Why the last two lines of section 3 (using
Equals()
) both return false? Why is there difference to section 1 line 3? Why isn’t the overload and the value comparison made? It is getting quite abstract and I can’t find the reason.
回答1:
In section 1 line 3
int a = 0;
short b = 0;
Console.WriteLine(a.Equals(b));
you call this overload of int
: bool Equals(int other)
, because b
(short) can be implicitly converted to int
so this overload is chosen. It returns true. In Section 3 line 3
int a = 0;
short b = 0;
object a1 = a;
object b1 = b;
Console.WriteLine(a1.Equals(b1)); //???
another overload of int
(not object
, because Equals
is virtual method) is called: bool Equals(object other)
. For it to return true other
should have exactly the same type (int
), but it's really short
so it returns false. Boxing is not relevant here, which you can verify with this:
int a = 0;
int c = 0;
object a1 = a;
object c1 = c;
// yes, different objects
Console.WriteLine(a1 == c1); // false
// still equal, because both are boxed ints
Console.WriteLine(a1.Equals(c1)); // true
As for understanding, I think documentation contains all relevant information. Just remember that:
Both
==
operator andEquals
method can be manually defined in class and so in theory can do anything. Your understanding relates only to "default" behavior.==
is not virtual in common sense, unlikeEquals
method. So when you doa1 == b1
-==
to be called in defined in compile time (based on types ofa1
andb1
), but when you calla1.Equals(b1)
- it is virtually dispatched so method to call is defined at runtime.
回答2:
In addition to boxing which means that a value type will result in a different reference in memory you have to consider Implicit Numeric Conversions in fact, for this reason, you have
Console.WriteLine(a.Equals(b));
which gives you a true but not this
Console.WriteLine(b.Equals(a));
here another example
static void Main(string[] args)
{
int i = 0;
long L = 0;
Console.WriteLine(1);
//comparing primitives - int vs. short
Console.WriteLine(L.Equals(i)); //true
Console.WriteLine(i.Equals(L));//false
}
来源:https://stackoverflow.com/questions/46829332/confusion-about-comparison-by-equals-vs-operator-and-primitives-vs-objec