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Re: double-to-int conversion without a cast

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  Re: double-to-int conversion without a cast Roedy Green <inquiry@mindprod.com> - 2016-06-02 22:18 -0700
    Re: double-to-int conversion without a cast Martin Gregorie <martin@address-in-sig.invalid> - 2016-06-03 10:53 +0000
      Re: double-to-int conversion without a cast Eric Sosman <esosman@comcast-dot-net.invalid> - 2016-06-03 08:22 -0400
        Re: double-to-int conversion without a cast bursejan@gmail.com - 2016-06-05 16:57 -0700

#34449 — Re: double-to-int conversion without a cast

On 25 Feb 2016 01:07:44 GMT, ram@zedat.fu-berlin.de (Stefan Ram)
wrote:

>ouble to int
>  with only the above means

I asked the conversion amanusensis at
http://mindprod.com/applet/converter.html

it offered:

// to int i from double d
// best
i = (int)d;
// or
i = (int)Math.round(d);
// or
i = (int)Math.ceil(d);
// or
i = (int)Math.floor(d);

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#34452

On Thu, 02 Jun 2016 22:18:13 -0700, Roedy Green wrote:

> On 25 Feb 2016 01:07:44 GMT, ram@zedat.fu-berlin.de (Stefan Ram)
> wrote:
> 
>>ouble to int
>>  with only the above means
> 
> I asked the conversion amanusensis at
> http://mindprod.com/applet/converter.html
> 
> it offered:
> 
> // to int i from double d // best i = (int)d;
> // or i = (int)Math.round(d);
> // or i = (int)Math.ceil(d);
> // or i = (int)Math.floor(d);

This works:

 int  n = new Long(Math.round(d)).intValue();

Look Ma, no casts!


-- 
martin@   | Martin Gregorie
gregorie. | Essex, UK
org       |

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#34453

On 6/3/2016 6:53 AM, Martin Gregorie wrote:
> On Thu, 02 Jun 2016 22:18:13 -0700, Roedy Green wrote:
>
>> On 25 Feb 2016 01:07:44 GMT, ram@zedat.fu-berlin.de (Stefan Ram)
>> wrote:
>>
>>> ouble to int
>>>  with only the above means
>>
>> I asked the conversion amanusensis at
>> http://mindprod.com/applet/converter.html
>>
>> it offered:
>>
>> // to int i from double d // best i = (int)d;
>> // or i = (int)Math.round(d);
>> // or i = (int)Math.ceil(d);
>> // or i = (int)Math.floor(d);
>
> This works:
>
>  int  n = new Long(Math.round(d)).intValue();
>
> Look Ma, no casts!

     Too expensive -- just *look* at that wasteful creation of an object!
Tut, tut, tut!  Try this:

       int n = Integer.parseInt(
             Double.toString(d)
                .substring(0, Double.toString(d).indexOf('.')))
             + (Double.toString(d).indexOf('.') >= 0
                && Double.toString(d).indexOf('.') + 1
                    < Double.toString(d).length()
                && Double.toString(d).charAt(
                   Double.toString(d).indexOf('.') + 1) >= '5'
             ? Double.toString(d).startsWith("-") ? -1 : 1 : 0);

     Look, Ma, no `new'!

-- 
esosman@comcast-dot-net.invalid
"Don't be afraid of work. Make work afraid of you." -- TLM

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#34482

This code:

    public static void main(String[] args) {
        double d = Math.PI*100;
        int x = (int)d;
        System.out.println("x="+x);
    }

Prints:

    x=314

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Basically you get zero for NaN, min_integer for underflow
and max_integer for overflow.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

See also:
https://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.3
- double to byte, short, char, int, long, or float

If the floating-point number is NaN (§4.2.3), the result of the first step of the conversion is an int or long 0.

Otherwise, if the floating-point number is not an infinity, the floating-point value is rounded to an integer value V, rounding toward zero using IEEE 754 round-toward-zero mode (§4.2.3). Then there are two cases:

    If T is long, and this integer value can be represented as a long, then the result of the first step is the long value V.

    Otherwise, if this integer value can be represented as an int, then the result of the first step is the int value V.

Otherwise, one of the following two cases must be true:

    The value must be too small (a negative value of large magnitude or negative infinity), and the result of the first step is the smallest representable value of type int or long.

    The value must be too large (a positive value of large magnitude or positive infinity), and the result of the first step is the largest representable value of type int or long.

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