CODEWITHSUNDEEP
javacastingfloating-pointdoubleprecision
user827992
user827992

Reputation: 1753

Understanding Casting in Java

this is my first post here and i like to thank all the people that can help me with this simple question: how the casting works in java?

I made this very simple class:

public class test {
    public static void main ( String[] args )
    {
        System.out.println((short)(1/3));
        System.out.println((int)(1/3));
        System.out.println((float)(1/3));
        System.out.println((double)(1/3));
    }
}

and this piece of software gives me this output when executed ( official JDK 6 u26 on a 32 bit machine under linux )

0
0
0.0
0.0

the problem, or the thing that i don't understand if you would, is that the last 2 results are 0.0, i was expecting something like 0.3333333, but apparently the cast works in another way: how?

thanks

PS I'm not so familiar with the english language, if i made some errors i apologize for this

Upvotes: 4

Views: 437

Answers (6)

Will
Will

Reputation: 20235

Because you have the 1/3 in parentheses, it uses integer division then casts to the type specified. It would evaluate to .3333333 if you used 1.0/3.0 or (float)1/3. 1 and 3 are integer literals, so results from division will be truncated. To use float and double literals put an f or d at the end of the number (1.0f, 1.0d). Casting affects the value after it, so if your value is in parentheses, it will evaluate that, then make the cast.

Upvotes: 1

Sean Patrick Floyd
Sean Patrick Floyd

Reputation: 299218

You need to cast the operands, not the result. This will yield the output you expected:

System.out.println(((short) 1 / (short) 3));
System.out.println((1 / 3));
System.out.println(((float) 1 / (float) 3));
System.out.println(((double) 1 / (double) 3));

Upvotes: 1

Umesh Kacha
Umesh Kacha

Reputation: 13686

int is coverted into double.

(1/3) is of type int so answer will be 0.

Now 0 is casted to double or float so it has become 0.0

Upvotes: 1

JB Nizet
JB Nizet

Reputation: 692181

First, the expression 1/3 is executed. This expression means "integer division of 1 by 3". And its result is the integer 0. Then this result (0) is cast to a double (or float). And of course it leads to 0.0.

You must cast one of the operands of the division to a double (or float) to transform it into a double division:

double d = ((double) 1) / 3;

or simply write

1.0 / 3

Upvotes: 9

Alex
Alex

Reputation: 365

In the last 2 cases you first compute 1/3, which is 0. Then you cast it to a float/double. Use:

((float)1)/3

Upvotes: 4

Jean Hominal
Jean Hominal

Reputation: 16796

If you use the operator / on integer values, the result will be typed as an integer.

You will have to force either 1 or 3 to a floating point value - try writing 1.0/3 instead of 1/3.

Upvotes: 5

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