Can I define a method that only accepts primitive types?

Question

I want to create a constructor that only accepts primitive types, how can I do this?

Like this example:

public Test(PrimitiveType type)
{

}

I need do it in a constructor and it's optional, so I want to create a parameterless constructor and a constructor with the parameter.

Answer 1

Depending upon what you want to achieve, you might want to look at so-called "convertible types", e.g. the types that implement IConvertible interface, those are the following:

• Boolean,
• SByte,
• Byte,
• Int16,
• UInt16,
• Int32,
• UInt32,
• Int64,
• UInt64,
• Single,
• Double,
• Decimal,
• DateTime,
• Char, and
• String.

So, as you see, this covers pretty much of what you'd like to achieve with primitive types.

So, by writing the method like that

public void Test(IConvertible primitive)
{
   if (primitive is Double) ....
   if (primitive is String) ....
}

you will confine your input types to the following ones (no structs etc).

Alternatively, you can also implement it as a generic method:

public void Test<T>(T primitive) where T : IConvertible
{
   if (primitive is Double) ....
   if (primitive is String) ....
}

Since you put this constrain, you can always convert your type to one, like:

public void Test<T>(T primitive) where T : IConvertible
{
   var myval = Convert.ToDecimal(primitive);
   ....
}

Answer 2

There is no way to do it with a single overload¹ (but you could write overloads for each primitive type, of course). You can make it accept only value types, but then it will accept any struct, not just primitive types.

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¹ well, it's not possible to enforce it at compile time, but you can check the type at runtime and throw an exception of course...