int promotion to unsigned int in C and C#

Question

Have a look at this C code:

int main()
{
    unsigned int y = 10;
    int x = -2;
    if (x > y)
        printf("x is greater");
    else
        printf("y is greater");
    return 0;
}
/*Output: x is greater.*/ 

I understand why the output is x is greater, because when the computer compares both of them, x is promoted to an unsigned integer type. When x is promoted to unsigned integer, -2 becomes 65534 which is definitely greater than 10.

But why in C#, does the equivalent code give the opposite result?

public static void Main(String[] args)
{
    uint y = 10;
    int x = -2;
    if (x > y)
    {
        Console.WriteLine("x is greater");
    }
    else
    {
        Console.WriteLine("y is greater");
    }
}
//Output: y is greater. 

Answer 1

In C#, both uint and int get promoted to a long before the comparison.

This is documented in 4.1.5 Integral types of the C# language spec:

For the binary +, –, *, /, %, &, ^, |, ==, !=, >, <, >=, and <= operators, the operands are converted to type T, where T is the first of int, uint, long, and ulong that can fully represent all possible values of both operands. The operation is then performed using the precision of type T, and the type of the result is T (or bool for the relational operators). It is not permitted for one operand to be of type long and the other to be of type ulong with the binary operators.

Since long is the first type that can fully represent all int and uint values, the variables are both converted to long, then compared.