CODEWITHSUNDEEP
javafloating-point
dev_coders
dev_coders

Reputation: 61

Why does the float default value is not 0.0f in java in the below code

 public class Main
    {
        float fValue;
        public static void main(String[] args) {
            Main obj=new Main();
            System.out.println(obj.fValue);
        }
    }

Here the output is 0.0 and not 0.0f.Can anyone explain it?

Upvotes: 1

Views: 1821

Answers (1)

Thor Odinson
Thor Odinson

Reputation: 46

That's because the "f" used after a float's digits is actually used in creation time so as to mark the value as a float. Eg. The number 123.45f is a float as it has been marked by an "f". The number 123.45d is a double, as it is marked by "d". Example, see the two cases below

Instantiating a float variable

float floatVariable = 123.45f;

Passing a float variable to a method.

method declaration:

void printFloat(float floatValue){
     System.out.println(floatValue);
}

method call:

printFloat(floatVariable);
printFloat(123.45f);

All values with decimal digits are by default double, and all numbers without decimal digits are by default integers. The "f" is required to declare the number as float.

printFloat((float)12345);

Above, you will have to cast to float because 12345 is an integer.

printFloat((float)123.45)

Above, you will have to cast to float because 123.45 is a double.

printFloat(123.45f);

Excellent!

At the other hand, printing the variable only prints the value. It is not a case of creation, so no necessary qualifier is needed.

Upvotes: 1

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