C#•5mo ago

✅ Two methods: struct and non-struct

Why does b.F(42) call the 2nd overload? 42 is a value-type which is what structs are?

public class Base
{
  public virtual void F<T>(T? t) where T : struct { }
  public virtual void F<T>(T? t) { }
}

Base b = new();
int? nullInt = null;
int? nonNullNullableInt = 42;

// These call the 1st overload.
b.F(nullInt);
b.F(nonNullNullableInt);
b.F(default(int?));

// These call the 2nd overload.
b.F(42);
b.F("Hello");
b.F(default(int));

public class Base
{
  public virtual void F<T>(T? t) where T : struct { }
  public virtual void F<T>(T? t) { }
}

Base b = new();
int? nullInt = null;
int? nonNullNullableInt = 42;

// These call the 1st overload.
b.F(nullInt);
b.F(nonNullNullableInt);
b.F(default(int?));

// These call the 2nd overload.
b.F(42);
b.F("Hello");
b.F(default(int));

6 Replies

reflectronic•5mo ago

because calling the first overload would require converting into a Nullable<T> (a.k.a. T? when T is a struct) while the second overload does not require any conversion, so it is a closer match

Jason_BjornOP•5mo ago

Thank you. If it did use the first one, would that involve boxing it?

reflectronic•5mo ago

no, it is not a boxing conversion it is just a wrapper struct struct Nullable<T> { T value; bool hasValue; } which is blessed by the language

Jason_BjornOP•5mo ago

thank you @reflectronic

Unknown User•5mo ago

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MODiX•5mo ago

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✅ Two methods: struct and non-struct