Console displaying char as an int when using a Conditional Operator ( ?: )

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I am surprised by the output of the following code

public static void Main(string[] args)
{
    char x = 'A';
    int i = 0;
    Console.WriteLine (true  ? x : 0);
    Console.WriteLine(false ? i : x); 
}

When I was reading the C# interview question I saw this code and the output of above code was

Output 

65
65

I am wondering how does that happen.

Can anybody explain me ? Thanks!

console-application c#
1 Answers

'A' is a char and has the value of 65

However, there is a technical explanation for why this results in an integer (and not the string representation of a char), you can find it in the ECMA C# specifications

12.15 Conditional operator

The second and third operands, x and y, of the ?: operator control the type of the conditional expression.

  • If x has type X and y has type Y then,
    • If X and Y are the same type, then this is the type of the conditional expression.
    • Otherwise, if an implicit conversion (§11.2) exists from X to Y, but not from Y to X, then Y is the type of the conditional expression.
    • Otherwise, if an implicit enumeration conversion (§11.2.4) exists from X to Y, then Y is the type of the conditional expression.
    • Otherwise, if an implicit enumeration conversion (§11.2.4) exists from Y to X, then X is the type of the conditional expression.
    • Otherwise, if an implicit conversion (§11.2) exists from Y to X, but not from X to Y, then X is the type of the conditional expression.
    • Otherwise, no expression type can be determined, and a compile-time error occurs.
  • If only one of x and y has a type, and both x and y are implicitly convertible to that type, then that is the type of the conditional expression.
  • Otherwise, no expression type can be determined, and a compile-time error occurs

Example

char Y = 'A';
int X = Y;
Y = X; // compiler error

In short, there is no implicit conversion from int to char, yet there is from char to int, so it makes sense for the result type to be an int

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