unsigned int (c++) vs uint (c#)

Question

Following is the c# code:

   static void Main(string[] args)
    {
        uint y = 12;
        int x = -2;
        if (x > y)
            Console.WriteLine(\         


        

Answer 1

I don't know about the standard of C#, but in the C++ standard, usual arithmetic conversions would be applied to both operands of relational operators:

[......enum, floating point type involed......] 

— Otherwise, the integral promotions (4.5) shall be performed on both operands.
  Then the following rules shall be applied to the promoted operands:

    — If both operands have the same type, no further conversion is needed.

    — Otherwise, if both operands have signed integer types or both have
      unsigned integer types, the operand with the type of lesser integer
      conversion rank shall be converted to the type of the operand with
      greater rank.

    — Otherwise, if the operand that has unsigned integer type has rank
      greater than or equal to the rank of the type of the other operand, the
      operand with signed integer type shall be converted to  the type of the
      operand with unsigned integer type.

    — Otherwise, if the type of the operand with signed integer type can
      represent all of the values of the type of the operand with unsigned
      integer type, the operand with unsigned integer type shall be converted
      to the type of the operand with signed integer type.

    — Otherwise, both operands shall be converted to the unsigned integer type 
      corresponding to the type of the operand with signed integer type.

Thus, when unsigned int is compared with int, int would be converted to unsigned int, and -2 would become a very large number when converted to unsigned int.