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Generic method difference between Int32 and int

I have problem with understanding generic method in interface. I'm using .NET framework 4.8 with auto C# version so it should be C# 7.3. Example below

Interface is:

public interface IRegister
{
  Nullable<T> Read<T>() where T: struct;
  void Write<T>(T value) where T : struct;
}

Now some class inherit from interface:

public abstract class AbstractRegister : IRegister
{
  protected ModbusClient client;
  protected int Address;
  public abstract Nullable<T> Read<T>() where T : struct;
  public virtual void Write<T>(T value) where T : struct
  {
    return;
  }
  public AbstractRegister(int address, ModbusClient client)
  {
    this.client = client;
    this.Address = address;
  }
}

Now I have some class inherited from AbstractRegister:

class InputRegister_int : AbstractRegister
{
  public InputRegister_int(int address, ModbusClient client) : base(address, client) { }

  public override int? Read<int>()
  {
    if (client == null)
      return null;
    return this.client.ReadInputRegisters(this.Address, 1)[0] as int?;
  }
}

and in line public override int? Read<int>() I have errors:

CS1001: Identifier expected
CS1003: Syntax error, '>' expected
CS1003: Syntax error, '(' expected

I'm using Visual Studio Community 2022

When I use Int32 instead int, everything works ok, and it almost resolve my problem, but type float (which is necessary for me) doesn't have Float representation.

I have no ideas for solving this problem anymore. Can someone explain it to me?

over 4 years ago · Santiago Trujillo
1 Respuestas
Responde la pregunta

0

This answer is specifically about "Int32" vs. int part of the question.

Int32 used in the code as example of "this works" is not actually System.Int32 (which is exactly what int is) but rather just a name that happen to match name of some system type.

Code below shows simpler example of this - you can see that "Int32" behaves exactly as any other string that is not a type name.

using System;
public class SomeClass {

    // exactly the same as T X<T>(), just using `Int32` as type parameter name
    Int32 X<Int32>() { return default(Int32); } 
    
    // fails to compile as `System.Int32` and `int` are concrete types
    System.Int32 Y<System.Int32>() { return default(System.Int32); } 
    int X<int>() { return default(int); } 
}

Conventionally name of the type parameter in generic method starts with T (just T or prefixed with it - TResult), but there is rule in C# specification that enforces that. As result anything that is not existing type name is considered "name of the type parameter" as long as it is used in the right place void M<AnythingThatIsNotType>(...) .

over 4 years ago · Santiago Trujillo Denunciar
Responde la pregunta
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