Int64 % Int32 gives Int64 result

Question

long longVar = 100_000_000_000;
int intVar = int.MinValue;
long result = longVar % intVar;

In this example, why result should be long? It cannot be more than Int32.MaxValue, why it was decided to make remainder Int64 in this operation?

Answer 1

As per the C# specification only the following remainder operators are predefined for integer types:

int operator %(int x, int y);
uint operator %(uint x, uint y);
long operator %(long x, long y);
ulong operator %(ulong x, ulong y);

Hence in your case the compiler chooses the long(long, long) version, and casts intVar to long automatically. Then the result is of type long .

Answer 2

Because if you try to use Int32 in this situation will be OverflowException(maybe). And have any moments in which will be cast into bigger object. For example:

int firstNumber = 10;
long secondNumber = 100;
var result = firstNumber + secondNumber;

(result - long) If you don't believe me you can check.

This is due to the automatic cast to the parameter type of a specific method. In your case, this is '%'. In the case of the '+' sign, you can override the statement and specify your own implementation of this statement.