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Why doesn't Math.Round/Floor/Ceiling return long or int?
msdn defined this method:Returns the smallest integer greater than or equal to the specified double-precision floating-point number.
but in fact,it is
public static double Ceiling ( double a ) why not return int directly? what does microsoft think of ?
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The documentation says about the return value: The smallest whole number greater than or equal to a. If a is equal to NaN, NegativeInfinity, or PositiveInfinity, that value is returned. Therefore the return value has to be double since NaN, NegativeInfinity and PositiveInfinity are fields of Double.
The Math.ceil() function always rounds up and returns the smaller integer greater than or equal to a given number.
ceil() returns the double value that is greater than or equal to the argument and is equal to the nearest mathematical integer. Note: If the argument is Integer, then the result is Integer. If the argument is NaN or an infinity or positive zero or negative zero, then the result is the same as the argument.
The Math. Ceiling() method in C# is used to return the smallest integral value greater than or equal to the specified number.
It's because the range of a double (±5.0 × 10−324 to ±1.7 × 10308) is much greater than the range of an int (-2,147,483,648 to 2,147,483,647). If the return type were int many possible inputs would fail. For example Math.Ceiling might be forced to throw an OverflowException in a checked context, or it might even return an incorrect result in an unchecked context. This is undesirable behaviour.
Also some special values such as NaN and PositiveInfinity can be returned by this method. This is only possible if the return type is double.
If you believe that the result will fit into an int, you can add an explicit cast:
int result = (int)Math.Ceiling(a); If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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