[FIXED] Why can I pass 1 as a short, but not the int variable i?

Issue

Why does the first and second Write work but not the last? Is there a way I can allow all 3 of them and detect if it was 1, (int)1 or i passed in? And really why is one allowed but the last? The second being allowed but not the last really blows my mind.

Demo to show compile error

using System;
class Program
{
    public static void Write(short v) { }
    static void Main(string[] args)
    {
        Write(1);//ok
        Write((int)1);//ok
        int i=1;
        Write(i);//error!?
    }
}

Solution

The first two are constant expressions, the last one isn't.

The C# specification allows an implicit conversion from int to short for constants, but not for other expressions. This is a reasonable rule, since for constants the compiler can ensure that the value fits into the target type, but it can't for normal expressions.

This rule is in line with the guideline that implicit conversions should be lossless.

6.1.8 Implicit constant expression conversions

An implicit constant expression conversion permits the following conversions:

• A constant-expression (§7.18) of type int can be converted to type sbyte, byte, short, ushort, uint, or ulong, provided the value of the constant-expression is within the range of the destination type.
• A constant-expression of type long can be converted to type ulong, provided the value of the constant-expression is not negative.

(Quoted from C# Language Specification Version 3.0)

Answered By - CodesInChaos

Answer Checked By - Clifford M. (PHPFixing Volunteer)