问题
private static Matcher<T> EqualTo<T>(T item)
{
return new IsEqual<T>(item);
}
How do I modify the above method definition such that the following are valid/allowed.
EqualTo("abc");
EqualTo(4);
EqualTo(null); // doesn't compile. EqualTo<string>(null) does
Trying to port some Java code where null seems to be acceptable value for a T parameter.
Update
Thanks: for all the answers - especially Eamon and Jason. I didn't want the method calls to bother with type-inference. The following overload fixed it.
private static Matcher<object> EqualTo(object item)
{
return EqualTo<object>(item);
}
Actually the above question was a part of a larger puzzle. The end goal was for the following to work.
this.AssertThat(null, EqualTo(null));
this.AssertThat(null, Not(EqualTo("hi")));
this.AssertThat("hi", Not(EqualTo(null)));
Applied the same fix.. RFC. (Ignore the ugly extension method part - that's another problem. Wanted to have these methods in all test-fixtures without inheritance.)
public static void AssertThat<T>(this object testFixture, object actual, Matcher<T> matcher, string message = "")
{
AssertThat(anyObject, (T)actual, matcher, message);
}
public static void AssertThat<T, TSuper>(this object testFixture, T actual, Matcher<TSuper> matcher, string message = "") where T : TSuper
{
... check and assert
回答1:
Consider the following method:
public bool IsNullString<T>(T item) {
return typeof(T) == typeof(string) && item == null;
}
Yes, this is a pathetically stupid method and using generics is pointless here, but you'll see the point in a moment.
Now consider
bool first = IsNullString<string>(null);
bool second = IsNullString<Foo>(null);
bool third = IsNullString(null);
In the first and second, the compiler can clearly distinguish the type of T (no inference is needed). In the third, how the compiler infer what T is? In particular, it can't distinguish between T == string and T == Foo, or any other type for that matter. Therefore, the compiler has to give you a compile-time error.
If you want to get around this, you either need to cast null
EqualTo((object)null);
or explicitly state the type
EqualTo<object>(null)
or define an overload
private static Matcher<object> EqualTo(object item) {
return new IsEqual<object>(item);
}
回答2:
Not possible without explicitly specifying a T or doing a cast. Generics are compile time constructs and as such if the compiler can't figure out the type at compile time, then it won't compile (as you're seeing).
回答3:
Since you can't do exactly what you are wanting to do, how about defining an EqualTo(object) overloaded method? That should allow your required syntax.
回答4:
You may work around this limitation by using the following syntax:
EqualTo("abc");
EqualTo(4);
EqualTo(default(object));
//equivalently:
EqualTo((object)null);
default(T) is the value a field of type T has if not set. For reference types, it's null, for value types it's essentially memory filled with zero bytes (...which may mean different things for different types, but generally means some version of zero).
I try to avoid the null everywhere in my code nowadays. It hampers type inference elsewhere too, such as with the var declared field and in a ternary operator. For example, myArray==null ? default(int?) : myArray.Length is OK, but myArray==null ? null : myArray.Length won't compile.
回答5:
Maybe implementing a non-generic EqualTo, which takes an Object as the argument type, would solve the issue of rewriting those code lines.
来源:https://stackoverflow.com/questions/5612489/how-to-allow-a-generic-type-parameter-for-a-c-sharp-method-to-accept-a-null-argu