Double to int and double to int

Question

I want to know what happened when Double invert to int on my example?

this case is right

Double d = 123.13;
int i = (int)(Math.random()*d);

this case is wrong

Double d = 123.13;
int i = (int)(d);

error message:Inconvertible types;cannot cast'java.lang.Double'to'int'

and I know I can use d.intValue to do this, but I just want to know what is the difference about the two situations above

`Double` cannot be cast to `int`: because `int` is not a sub-type of `Double` — ernest_k, Nov 07 '19 at 07:15
@ernest_k `double` isn't a sub type of `Double` either. It's just that the conversion is defined. — Andy Turner, Nov 07 '19 at 07:22
@AndyTurner Right. When there's no special relationship between the two, we check super/sub-type relationships. That's where I came from. *But I know I'm now taking this too far :-)* — ernest_k, Nov 07 '19 at 07:32
double is `primitive` but Double is `Class` , you know `d.intValue` actual cast `double` to `int`, but be careful not cast `Double` to `int`. — huang botao, Nov 07 '19 at 07:49

score 3 · Answer 1 · answered Nov 07 '19 at 07:15

3

A wrapper type can be cast only to the primitive type that it wraps.

i.e. you can cast a Double to a double, but not directly to an int.

The second snippet can pass compilation with:

Double d = 123.13;
int i = (int)(double)(d);

In the first snippet, you are casting a primitive (the type of Math.random()*d is double), to an int, which is allowed.

answered Nov 07 '19 at 07:15

Eran

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thanks for your answer what is very detailedness!! – achilles0009 Nov 07 '19 at 07:20

score 0 · Answer 2 · answered Nov 07 '19 at 07:19

0

You could cast the primitive value like this, too:

Double d = 123.13;
int i = (int) d.doubleValue();

…which is pretty much the same as the answer given by Eran.

answered Nov 07 '19 at 07:19

deHaar

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Yes, or `int i = d.intValue();`. – Maurice Perry Nov 07 '19 at 07:22
nice,thank you so much – achilles0009 Nov 07 '19 at 07:33

score 0 · Answer 3 · answered Nov 07 '19 at 07:23

Situation 1: Double d = 123.13;int i = (int)(Math.random()*d);

Situation 2: Double d = 123.13;int i = (int)(d);

What is difference about the two situations above?

In the first case Math.random() returns a double, not Double, so this one is right because double can be cast to int (losing the fraction part).

Situation 2 is wrong and the reason is explained in @Eran's answer, A wrapper type can be cast only to the primitive type that it wraps. i.e you cannot cast Double to nonconvertible primitive int

1

understand,thank you – achilles0009 Nov 07 '19 at 07:34

score 0 · Answer 4 · edited Jun 20 '20 at 09:12

There is lots of thing going on, see JLS 5.1.7. Boxing Conversion .

2nd one is wrong because Double is class and double is primitive data type

Narrowing is possible using primitive data type only

JLS 5.1.3. Narrowing Primitive Conversion

22 specific conversions on primitive types are called the narrowing primitive conversions:

short to byte or char

char to byte or short

int to byte, short, or char

long to byte, short, char, or int

float to byte, short, char, int, or long

double to byte, short, char, int, long, or float

So to your code to work you can change your code like below

double d = 123.13;
int i = (int)d;

ok,i understand what you want to say,i'll go to see the Boxing Conversion .Thanks for your answer! — achilles0009, Nov 07 '19 at 07:32

Double to int and double to int

4 Answers4